MathDB
1987 AJHSME Problem 25

Source:

June 25, 2011
probability

Problem Statement

Ten balls numbered 11 to 1010 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
(A) 49(B) 919(C) 12(D) 1019(E) 59\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}
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