1984 AHSME Problems/Problem 19

Problem

A box contains $11$ balls, numbered $1, 2, 3, ... 11$ . If $6$ balls are drawn simultaneously at random, what is the probability that the sum of the numbers on the balls drawn is odd?

$\mathrm{(A) \ }\frac{100}{231} \qquad \mathrm{(B) \ }\frac{115}{231} \qquad \mathrm{(C) \ } \frac{1}{2} \qquad \mathrm{(D) \ }\frac{118}{231} \qquad \mathrm{(E) \ } \frac{6}{11}$

Solution

There are exactly $3$ ways the sum can be odd: exactly $5$ are odd, exactly $3$ are odd, or exactly $1$ is odd. There are exactly $6$ odd numbers in the set. For the first case, there are ${6\choose5}\times{5\choose1}=30$ possibilities. For the second case, there are ${6\choose3}\times{5\choose3}=200$ possibilities. For the third case, there are ${6\choose1}\times{5\choose5}=6$ possibilities. In each case, we're choosing out of the $6$ odd numbers which ones will be picked, and multiplying that by out of the $5$ even numbers, which one of those will be picked so that there are $6$ numbers. In total,there are $236$ desirable possibilities out of a total of ${11\choose6}=462$ possibilities. Thus, the answer is $\frac{236}{462}=\frac{118}{231}, \boxed{\text{D}}$ .

1984 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1984 AHSME Problems/Problem 19

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