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Difference between revisions of "2011 AMC 10A Problems/Problem 21"
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Note that we are trying to find the conditional probability <math>P(A \vert B) = \frac{P(A \cap B)}{P(B)}</math> where <math>A</math> is the <math>4</math> coins being genuine and <math>B</math> is the sum of the weight of the coins being equal. The only possibilities for <math>B</math> are (g is abbreviated for genuine and c is abbreviated for counterfeit, and the pairs are the first and last two in the quadruple) <math>(g,g,g,g),(g,c,g,c),(g,c,c,g),(c,g,g,c),(c,g,c,g)</math>. We see that <math>A \cap B</math> happens with probability <math>\frac{8}{10} \times \frac{7}{9} \times \frac{6}{8} \times \frac{5}{7} = \frac{1}{3}</math>, and <math>B</math> happens with probability <math>\frac{1}{3}+4 \times \left( \frac{8}{10} \times \frac{2}{9} \times \frac{7}{8}\times \frac{1}{7}\right) = \frac{19}{45}</math>, hence <math>P(A \vert B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{3}}{\frac{19}{45}}=\boxed{\frac{15}{19} \ \mathbf{(D)}}</math>.
 
Note that we are trying to find the conditional probability <math>P(A \vert B) = \frac{P(A \cap B)}{P(B)}</math> where <math>A</math> is the <math>4</math> coins being genuine and <math>B</math> is the sum of the weight of the coins being equal. The only possibilities for <math>B</math> are (g is abbreviated for genuine and c is abbreviated for counterfeit, and the pairs are the first and last two in the quadruple) <math>(g,g,g,g),(g,c,g,c),(g,c,c,g),(c,g,g,c),(c,g,c,g)</math>. We see that <math>A \cap B</math> happens with probability <math>\frac{8}{10} \times \frac{7}{9} \times \frac{6}{8} \times \frac{5}{7} = \frac{1}{3}</math>, and <math>B</math> happens with probability <math>\frac{1}{3}+4 \times \left( \frac{8}{10} \times \frac{2}{9} \times \frac{7}{8}\times \frac{1}{7}\right) = \frac{19}{45}</math>, hence <math>P(A \vert B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{3}}{\frac{19}{45}}=\boxed{\frac{15}{19} \ \mathbf{(D)}}</math>.
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== See Also ==
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{{AMC10 box|year=2011|ab=A|num-b=21|num-a=22}}

Revision as of 09:45, 8 May 2011

Problem 21

Two counterfeit coins of equal weight are mixed with $8$ identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the $10$ coins. A second pair is selected at random without replacement from the remaining $8$ coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all $4$ selected coins are genuine?

$\textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16}$

Solution

Note that we are trying to find the conditional probability $P(A \vert B) = \frac{P(A \cap B)}{P(B)}$ where $A$ is the $4$ coins being genuine and $B$ is the sum of the weight of the coins being equal. The only possibilities for $B$ are (g is abbreviated for genuine and c is abbreviated for counterfeit, and the pairs are the first and last two in the quadruple) $(g,g,g,g),(g,c,g,c),(g,c,c,g),(c,g,g,c),(c,g,c,g)$. We see that $A \cap B$ happens with probability $\frac{8}{10} \times \frac{7}{9} \times \frac{6}{8} \times \frac{5}{7} = \frac{1}{3}$, and $B$ happens with probability $\frac{1}{3}+4 \times \left( \frac{8}{10} \times \frac{2}{9} \times \frac{7}{8}\times \frac{1}{7}\right) = \frac{19}{45}$, hence $P(A \vert B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{3}}{\frac{19}{45}}=\boxed{\frac{15}{19} \ \mathbf{(D)}}$.

See Also

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 22
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All AMC 10 Problems and Solutions
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