Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2011 AMC 10A Problems/Problem 21

Revision as of 19:43, 18 February 2012 by Peace153 (talk | contribs)

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 1

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 $P(A \vert B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{3}}{\frac{19}{45}}=\boxed{\frac{15}{19} \ \mathbf{(D)}}$.

Solution 2

If we pick $4$ distinguishable real coins from the set of $8$ real coins, there are $\binom{8}{4}$ ways to pick the coins. If we then place the coins in four distinguishable slots on the scale, there are $4!$ ways to arrange them, giving $4!\cdot \binom{8}{4}$ ways to choose and place $8$ real coins. This gives $1680$ desirable combinations.

If we pick $2$ real coins and $2$ fake coins, there are $\binom{8}{2}\binom{2}{2}$ ways to choose the coins. There are $4$ choices for the first slot on the left side of the scale. Whichever type of coin is placed in that first slot, there are $2$ choices for the second slot on the left side of the scale, since it must be of the opposite type of coin. There are $2$ choices for the first slot on the right side of the scale, and only $1$ choice for the last slot on the right side.

Thus, there are $4\cdot 2\cdot 2\cdot 1 = 16$ ways to arrange the coins, and $\binom{8}{2}\binom{2}{2} = 28$ sets of possible coins, for a total of $16\cdot 28 = 448$ combinations that are legal, yet undesirable.

The overall probability is thus $\frac{1680}{1680 + 448} = \boxed{\frac{15}{19} \ \mathbf{(D)}}$.

Note that in this solution, all four slots on the scale are deemed to be distinguishable. In essense, this "overcounts" all numbers by a factor of $8$, since you can switch the coins on the left side, you can switch the coins on the right side, or you can switch sides of the scale. But since all numbers are increased 8-fold, it will cancel out when calculating the probability.

See Also

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_21&oldid=44827"