1987 AJHSME Problems/Problem 25

Problem

Ten balls numbered $1$ to $10$ are in a jar. Jack reaches into the jar and randomly removes one of the balls. Then Jill reaches into the jar and randomly removes a different ball. The probability that the sum of the two numbers on the balls removed is even is

$\text{(A)}\ \frac{4}{9} \qquad \text{(B)}\ \frac{9}{19} \qquad \text{(C)}\ \frac{1}{2} \qquad \text{(D)}\ \frac{10}{19} \qquad \text{(E)}\ \frac{5}{9}$

Solution 1

For the sum of the two numbers removed to be even, they must be of the same parity. There are five even values and five odd values.

No matter what Jack chooses, the number of numbers with the same parity is four. There are nine numbers total, so the probability Jill chooses a number with the same parity as Jack's is $\frac49$

$\boxed{\text{A}}$

1987 AJHSME (Problems • Answer Key • Resources)
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1987 AJHSME Problems/Problem 25

Problem

Solution 1

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