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Difference between revisions of "2000 AIME I Problems/Problem 5"
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<math>(3 + w_1)(22 - w_1) = 50</math>
 
<math>(3 + w_1)(22 - w_1) = 50</math>
  
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Since the factors of <math>50</math> that are greater than <math>3</math> are <math>5, 10, 25,</math> and <math>50</math>, the quantity <math>3 + w_1</math> must equal one of those.  However, testing <math>2, 7, 22</math> and <math>47</math> for <math>w_1</math> does not give a correct product.  Thus, <math>\frac{27}{50}</math> must be a reduced form of the actual fraction.
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Since the factors of <math>50</math> that are greater than <math>3</math> are <math>5, 10, 25,</math> and <math>50</math>, the quantity <math>3 + w_1</math> must equal one of those.  However, since <math>w_1 < 13</math>, testing <math>2</math> and <math>7</math> for <math>w_1</math> does not give a correct product.  Thus, <math>\frac{27}{50}</math> must be a reduced form of the actual fraction.
  
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First assume that the fraction was reduced from <math>\frac{54}{100}</math>, yielding the equations <math>b_1\cdot b_2 = 54</math> and <math>(b_1 + w_1)(b_2 + w_2) = 100</math>.
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Trying <math>b_1 \cdot b_2 = 54</math> with <math>b_1 <  b_2 < 25</math> gives <math>(b_1, b_2) = (3, 18)</math> or <math> (6, 9)</math>.  Trying the first pair and setting the denominator equal to 100 gives:
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Factoring <math>b_1 \cdot b_2 = 54</math> and saying WLOG that <math>b_1 <  b_2 < 25</math> gives <math>(b_1, b_2) = (3, 18)</math> or <math> (6, 9)</math>.  Trying the first pair and setting the denominator equal to 100 gives:
 
<math>(3 + w_1)(18 + w_2) = 100</math>
 
<math>(3 + w_1)(18 + w_2) = 100</math>
  
Line 48: Line 48:
  
  
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For completeness, the fraction <math>\frac{81}{150}</math> may be tested.  <math>150</math> is the highest necessary denominator that needs to be tested, since the maximum the denominator <math>(w_1+ b_1)(w_2 + b_2)</math> can be when the sum of all variables is <math>25</math> is when the variables are <math>6, 6, 6, </math> and <math>7</math>, in some permutation, which gives <math>154</math>.  If <math>b_1 \cdot b_2 = 81</math>, this forces <math>b_1 = b_2 = 9</math>, since all variables must be integers under <math>25</math>.  The denominator becomes <math>(9 + w_1)(9 + w_2) = 150</math>, and since there are now <math>25 - 18 = 7</math> white marbles total, the denominator becomes <math>(9 + w_1)(16 - w_1) = 150</math>.  Testing <math>w_1 = 1</math> gives a solution, and thus <math>w_2 = 6</math>.  Although the distribution and colors of the marbles is different from the last case, the probability of drawing two white marbles is <math>\frac{6 \cdot 1}{ 150}</math>, which still simplifies to <math>\frac {1}{25}</math>.
+
For completeness, the fraction <math>\frac{81}{150}</math> may be tested.  <math>150</math> is the highest necessary denominator that needs to be tested, since the maximum the denominator <math>(w_1+ b_1)(w_2 + b_2)</math> can be when the sum of all integer variables is <math>25</math> is when the variables are <math>6, 6, 6, </math> and <math>7</math>, in some permutation, which gives <math>154</math>.  If <math>b_1 \cdot b_2 = 81</math>, this forces <math>b_1 = b_2 = 9</math>, since all variables must be integers under <math>25</math>.  The denominator becomes <math>(9 + w_1)(9 + w_2) = 150</math>, and since there are now <math>25 - 18 = 7</math> white marbles total, the denominator becomes <math>(9 + w_1)(16 - w_1) = 150</math>.  Testing <math>w_1 = 1</math> gives a solution, and thus <math>w_2 = 6</math>.  The complete solution for this case is <math>(w_1, w_2, b_1, b_2) = (1, 6, 9, 9)</math>.  Although the distribution and colors of the marbles is different from the last case, the probability of drawing two white marbles is <math>\frac{6 \cdot 1}{ 150}</math>, which still simplifies to <math>\frac {1}{25}</math>.
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== See also ==
 
== See also ==

Revision as of 15:45, 23 July 2011

Problem

Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $27/50,$ and the probability that both marbles are white is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m + n$?

Solution 1

If we work with the problem for a little bit, we quickly see that there is no direct combinatorics way to calculate $m/n$. The Principle of Inclusion-Exclusion still requires us to find the individual probability of each box.

Let $a, b$ represent the number of marbles in each box, and without loss of generality let $a>b$. Then, $a + b = 25$, and since the $ab$ may be reduced to form $50$ on the denominator of $\frac{27}{50}$, $50|ab$. It follows that $5|a,b$, so there are 2 pairs of $a$ and $b: (20,5),(15,10)$.

  • Case 1: Then the product of the number of black marbles in each box is $54$, so the only combination that works is $18$ black in first box, and $3$ black in second. Then, $P(\text{both white}) = \frac{2}{20} \cdot \frac{2}{5} = \frac{1}{25},$ so $m + n = 26$.
  • Case 2: The only combination that works is 9 black in both. Thus, $P(\text{both white}) = \frac{1}{10}\cdot \frac{6}{15} = \frac{1}{25}$. $m + n = 26$.

Thus, $m + n = \boxed{026}$.

Solution 2

Let $w_1, w_2, b_1,$ and $b_2$ represent the white and black marbles in boxes 1 and 2.

Since there are $25$ marbles in the box:

$w_1 + w_2 + b_1 + b_2 = 25$

From the fact that there is a $\frac{27}{50}$ chance of drawing one black marble from each box:

$\frac{b_1 \cdot b_2}{(b_1 + w_1)(b_2 + w_2)} = \frac{27}{50} = \frac{54}{100} = \frac{81}{150}$

Thinking of the numerator and denominator separately, if $\frac{27}{50}$ was not a reduced fraction when calculating out the probability, then $b_1 \cdot b_2 = 27$. Since $b_1 < 25$, this forces the variables to be $3$ and $9$ in some permutation. Without loss of generality, let $b_1 = 3$ and $b_2 = 9$.

The denominator becomes: $(3 + w_1)(9 + w_2) = 50$

Since there have been $12$ black marbles used, there must be $13$ white marbles. Substituting that in:

$(3 + w_1)(9 + (13 - w_1)) = 50$

$(3 + w_1)(22 - w_1) = 50$

Since the factors of $50$ that are greater than $3$ are $5, 10, 25,$ and $50$, the quantity $3 + w_1$ must equal one of those. However, since $w_1 < 13$, testing $2$ and $7$ for $w_1$ does not give a correct product. Thus, $\frac{27}{50}$ must be a reduced form of the actual fraction.

First assume that the fraction was reduced from $\frac{54}{100}$, yielding the equations $b_1\cdot b_2 = 54$ and $(b_1 + w_1)(b_2 + w_2) = 100$. Factoring $b_1 \cdot b_2 = 54$ and saying WLOG that $b_1 < b_2 < 25$ gives $(b_1, b_2) = (3, 18)$ or $(6, 9)$. Trying the first pair and setting the denominator equal to 100 gives: $(3 + w_1)(18 + w_2) = 100$


Since $w_1 + w_2 = 4$, the pairs $(w_1, w_2) = (1, 3), (2,2),$ and $(3,1)$ can be tried, since each box must contain at least one white marble. Plugging in $w_1 = w_2 = 2$ gives the true equation $(3 + 2)(18 + 2) =100$, so the number of marbles are $(w_1, w_2, b_1, b_2) = (2, 2, 3, 18)$

Thus, the chance of drawing 2 white marbles is $\frac{w_1 \cdot w_2 }{(w_1+ b_1)(w_2 + b_2)} = \frac{4}{100} = \frac{1}{25}$ in lowest terms, and the answer to the problem is $1 + 25 = \boxed{026}.$


For completeness, the fraction $\frac{81}{150}$ may be tested. $150$ is the highest necessary denominator that needs to be tested, since the maximum the denominator $(w_1+ b_1)(w_2 + b_2)$ can be when the sum of all integer variables is $25$ is when the variables are $6, 6, 6,$ and $7$, in some permutation, which gives $154$. If $b_1 \cdot b_2 = 81$, this forces $b_1 = b_2 = 9$, since all variables must be integers under $25$. The denominator becomes $(9 + w_1)(9 + w_2) = 150$, and since there are now $25 - 18 = 7$ white marbles total, the denominator becomes $(9 + w_1)(16 - w_1) = 150$. Testing $w_1 = 1$ gives a solution, and thus $w_2 = 6$. The complete solution for this case is $(w_1, w_2, b_1, b_2) = (1, 6, 9, 9)$. Although the distribution and colors of the marbles is different from the last case, the probability of drawing two white marbles is $\frac{6 \cdot 1}{ 150}$, which still simplifies to $\frac {1}{25}$.

See also

2000 AIME I (Problems • Answer Key • Resources)
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Problem 6
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