Difference between revisions of "2009 AIME I Problems/Problem 3"

Revision as of 19:46, 19 March 2009

Problem

A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution

If we let the odds of a tails $(1-p)$ equal $t$ , then the probability of three heads and five tails is ${p^3}{t^5}$ The probability of five heads and three tails is ${p^5}{t^3}$

$25{p^3}{t^5} = {p^5}{t^3}$ $25{t^2} = {p^2}$ $5t = p$ $5(1 - p) = p$ $5 - 5p = p$ $5 = 6p$ $p = \frac {5} {6}$ $5 + 6 = \boxed{11}$

@@ Line 5: / Line 5: @@
 == Solution ==
-If we let the odds of a tails (1-p) equal t, then the probablity of three heads and five tails is:
+If we let the odds of a tails <math>(1-p)</math> equal <math>t</math>, then the probability of three heads and five tails is <math>{p^3}{t^5}</math>
-p^3t^5
+The probability of five heads and three tails is <math>{p^5}{t^3}</math>
-The probability of five heads and three tails is:
-p^5t^3
-p^3t^5 = p^5t^3
+<cmath>25{p^3}{t^5} = {p^5}{t^3}</cmath>
-t^2 = p^2
+<cmath>25{t^2} = {p^2}</cmath>
-t = p
+<cmath>5t = p</cmath>
-(1-p) = p
+<cmath>5(1 - p) = p</cmath>
-- 5p = p
+<cmath>5 - 5p = p</cmath>
-= 6p
+<cmath>5 = 6p</cmath>
-p = 5/6
+<cmath>p = \frac {5} {6}</cmath>
+<cmath>5 + 6 = \boxed{11}</cmath>

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Difference between revisions of "2009 AIME I Problems/Problem 3"

Revision as of 19:46, 19 March 2009

Problem

Solution