https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_29&feed=atom&action=history1992 AHSME Problems/Problem 29 - Revision history2024-03-28T11:10:50ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_29&diff=156521&oldid=prevTh1nq3r: /* Solution 2 */2021-06-20T23:29:11Z<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:29, 20 June 2021</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Sorry for the very hand-wavy and non-rigorous answer! At least mine has a cool recurrence relation! :D</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Sorry for the very hand-wavy and non-rigorous answer! At least mine has a cool recurrence relation! :D</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">(Or just come up with the general formula <math>P_n=\frac{1}{2}(1+\frac{1}{3^n})</math> Simple observation really. If you really want to get mathy, prove this through induction).</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>~th1nq3r</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>~th1nq3r</div></td></tr>
</table>Th1nq3rhttps://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_29&diff=156517&oldid=prevTh1nq3r: /* Solution 2 */2021-06-20T22:36:45Z<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 22:36, 20 June 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l15" >Line 15:</td>
<td colspan="2" class="diff-lineno">Line 15:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>\bold{Case <del class="diffchange diffchange-inline"> </del>1}</math>: (There are an odd number of heads for <math>(n-1)</math> coin flips).</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math>\bold{Case 1}</math>: (There are an odd number of heads for <math>(n-1)</math> coin flips).</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>\bold{Case <del class="diffchange diffchange-inline"> </del>2}</math>: (There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math>\bold{Case 2}</math>: (There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  </div></td></tr>
</table>Th1nq3rhttps://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_29&diff=156516&oldid=prevTh1nq3r: /* Solution 2 */2021-06-20T22:36:34Z<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 22:36, 20 June 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l15" >Line 15:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>\bold{Case 1}</math>: (There are an odd number of heads for <math>(n-1)</math> coin flips).</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math>\bold{Case <ins class="diffchange diffchange-inline"> </ins>1}</math>: (There are an odd number of heads for <math>(n-1)</math> coin flips).</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>\bold{Case 2}</math>: (There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math>\bold{Case <ins class="diffchange diffchange-inline"> </ins>2}</math>: (There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  </div></td></tr>
</table>Th1nq3rhttps://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_29&diff=156515&oldid=prevTh1nq3r: /* Solution 2 */2021-06-20T22:34:48Z<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 22:34, 20 June 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l15" >Line 15:</td>
<td colspan="2" class="diff-lineno">Line 15:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>\bold{Case 1}</math>:(There are an odd number of heads for <math>(n-1)</math> coin flips).</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math>\bold{Case 1}</math>: (There are an odd number of heads for <math>(n-1)</math> coin flips).</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>\bold{Case 2}</math> <del class="diffchange diffchange-inline">==</del>:(There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math>\bold{Case 2}</math>: (There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Sorry for the very hand-wavy and non-rigorous answer! At least mine has a cool recurrence relation! :D</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Sorry for the very hand-wavy and non-rigorous answer! At least mine has a cool recurrence relation! :D</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>~th1nq3r</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>~th1nq3r</div></td></tr>
</table>Th1nq3rhttps://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_29&diff=156514&oldid=prevTh1nq3r: /* Solution 2 */2021-06-20T22:34:27Z<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 22:34, 20 June 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l15" >Line 15:</td>
<td colspan="2" class="diff-lineno">Line 15:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">== </del>Case 1 <del class="diffchange diffchange-inline">==</del>:(There are an odd number of heads for <math>(n-1)</math> coin flips).</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline"><math>\bold{</ins>Case 1<ins class="diffchange diffchange-inline">}</math></ins>:(There are an odd number of heads for <math>(n-1)</math> coin flips).</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">== </del>Case 2 ==:(There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline"><math>\bold{</ins>Case 2<ins class="diffchange diffchange-inline">}</math> </ins>==:(There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  </div></td></tr>
</table>Th1nq3rhttps://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_29&diff=156513&oldid=prevTh1nq3r: /* Solution 2 */2021-06-20T22:33:49Z<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 22:33, 20 June 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l15" >Line 15:</td>
<td colspan="2" class="diff-lineno">Line 15:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>==Case 1==:(There are an odd number of heads for <math>(n-1)</math> coin flips).</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>== Case 1 ==:(There are an odd number of heads for <math>(n-1)</math> coin flips).</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>==Case 2==:(There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>== Case 2 ==:(There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  </div></td></tr>
</table>Th1nq3rhttps://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_29&diff=156512&oldid=prevTh1nq3r: /* Solution 2 */2021-06-20T22:33:35Z<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
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<col class="diff-content" />
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 22:33, 20 June 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l15" >Line 15:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline"><math></del>Case 1<del class="diffchange diffchange-inline"></math></del>:(There are an odd number of heads for <math>(n-1)</math> coin flips).</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">==</ins>Case 1<ins class="diffchange diffchange-inline">==</ins>:(There are an odd number of heads for <math>(n-1)</math> coin flips).</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline"><math></del>Case 2<del class="diffchange diffchange-inline"></math></del>:(There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">==</ins>Case 2<ins class="diffchange diffchange-inline">==</ins>:(There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed {D)}</math>.  </div></td></tr>
</table>Th1nq3rhttps://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_29&diff=156511&oldid=prevTh1nq3r: /* Solution 2 */2021-06-20T22:33:16Z<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 22:33, 20 June 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l20" >Line 20:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>Case 2</math>:(There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>Case 2</math>:(There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed D)</math>.  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed <ins class="diffchange diffchange-inline">{</ins>D)<ins class="diffchange diffchange-inline">}</ins></math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Sorry for the very hand-wavy and non-rigorous answer! At least mine has a cool recurrence relation! :D</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Sorry for the very hand-wavy and non-rigorous answer! At least mine has a cool recurrence relation! :D</div></td></tr>
</table>Th1nq3rhttps://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_29&diff=156510&oldid=prevTh1nq3r: /* Solution */2021-06-20T22:33:00Z<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 22:33, 20 June 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l11" >Line 11:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Doing casework on the number of heads (0 heads, 2 heads, 4 heads...), we get the equation <cmath>P=\left(\frac{1}{3} \right)^{50}+\binom{50}{2}\left(\frac{2}{3} \right)^{2}\left(\frac{1}{3} \right)^{48}+\dots+\left(\frac{2}{3} \right)^{50}</cmath> This is essentially the expansion of <math>\left(\frac{2}{3}+\frac{1}{3} \right)^{50}</math> but without the odd power terms. To get rid of the odd power terms in <math>\left(\frac{2}{3}+\frac{1}{3} \right)^{50}</math>, we add <math>\left(\frac{2}{3}-\frac{1}{3} \right)^{50}</math> and then divide by <math>2</math> because the even power terms that were not canceled were expressed twice. Thus, we have <cmath>P=\frac{1}{2}\cdot\left(\left(\frac{1}{3}+\frac{2}{3} \right)^{50}+\left(\frac{2}{3}-\frac{1}{3} \right)^{50} \right)</cmath> Or <cmath>\frac{1}{2}\left(1+\left(\frac{1}{3} \right)^{50} \right)</cmath> which is equivalent to answer choice <math>\fbox{D}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Doing casework on the number of heads (0 heads, 2 heads, 4 heads...), we get the equation <cmath>P=\left(\frac{1}{3} \right)^{50}+\binom{50}{2}\left(\frac{2}{3} \right)^{2}\left(\frac{1}{3} \right)^{48}+\dots+\left(\frac{2}{3} \right)^{50}</cmath> This is essentially the expansion of <math>\left(\frac{2}{3}+\frac{1}{3} \right)^{50}</math> but without the odd power terms. To get rid of the odd power terms in <math>\left(\frac{2}{3}+\frac{1}{3} \right)^{50}</math>, we add <math>\left(\frac{2}{3}-\frac{1}{3} \right)^{50}</math> and then divide by <math>2</math> because the even power terms that were not canceled were expressed twice. Thus, we have <cmath>P=\frac{1}{2}\cdot\left(\left(\frac{1}{3}+\frac{2}{3} \right)^{50}+\left(\frac{2}{3}-\frac{1}{3} \right)^{50} \right)</cmath> Or <cmath>\frac{1}{2}\left(1+\left(\frac{1}{3} \right)^{50} \right)</cmath> which is equivalent to answer choice <math>\fbox{D}</math>.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">== Solution 2 ==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"><math>Case 1</math>:(There are an odd number of heads for <math>(n-1)</math> coin flips).</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Well to solve this case, we first ask the question; "What is the probability we get an odd number of heads for <math>(n-1)</math> flips?" That is precisely <math>1-P_{n-1}</math>. Now, since we want an even number of heads, what is the probability that on the <math>n-th</math> coin flip we get a head? <math>\frac{2}{3}</math>. So we have <math>\frac{2}{3}(1-P_{n-1})</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"><math>Case 2</math>:(There are an even number of heads for <math>(n-1)</math> coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for <math>n</math> flips will be odd. :/ So what is the probability we get a tail? <math>1-\frac{2}{3}=\frac{1}{3}</math>. Thus we have <math>\frac{1}{3}(P_{n-1})</math>. </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Adding both cases, we have that <math>P_n=\frac{2}{3}-\frac{1}{3}P_{n-1}</math>. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in <math>\boxed D)</math>. </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Sorry for the very hand-wavy and non-rigorous answer! At least mine has a cool recurrence relation! :D</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">~th1nq3r</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
</table>Th1nq3rhttps://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_29&diff=107414&oldid=prevAopsuser101: /* Problem */2019-07-06T19:43:54Z<p><span dir="auto"><span class="autocomment">Problem</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 19:43, 6 July 2019</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Problem ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Problem ==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>An <del class="diffchange diffchange-inline">"</del>unfair<del class="diffchange diffchange-inline">" </del>coin has a <math>2/3</math> probability of turning up heads. If this coin is tossed <math>50</math> times, what is the probability that the total number of heads is even?</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>An unfair coin has a <math>2/3</math> probability of turning up heads. If this coin is tossed <math>50</math> times, what is the probability that the total number of heads is even?</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>\text{(A) } 25(\frac{2}{3})^{50}\quad</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math>\text{(A) } 25<ins class="diffchange diffchange-inline">\bigg</ins>(\frac{2}{3}<ins class="diffchange diffchange-inline">\bigg</ins>)^{50}\quad</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>\text{(B) } \frac{1}{2}(1-\frac{1}{3^{50}})\quad</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>\text{(B) } \frac{1}{2}<ins class="diffchange diffchange-inline">\bigg</ins>(1-\frac{1}{3^{50}}<ins class="diffchange diffchange-inline">\bigg</ins>)\quad</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\text{(C) } \frac{1}{2}\quad</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\text{(C) } \frac{1}{2}\quad</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>\text{(D) } \frac{1}{2}(1+\frac{1}{3^{50}})\quad</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>\text{(D) } \frac{1}{2}<ins class="diffchange diffchange-inline">\bigg</ins>(1+\frac{1}{3^{50}}<ins class="diffchange diffchange-inline">\bigg</ins>)\quad</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\text{(E) } \frac{2}{3}</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\text{(E) } \frac{2}{3}</math></div></td></tr>
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</table>Aopsuser101