1987 AJHSME Problems/Problem 25
Problem
Ten balls numbered to
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
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
Solution 2 - Complimentary Counting
An alternative approach would be to calculate the probability the sum of the two numbers is odd, and subtract that from the total. We see that there is only one way to get an odd sum, where the two numbers must be of opposite parity.
Here, we have two cases the two numbers are, respectively, (even, odd), and (odd, even). The probability to get an even number as the first number is
and an odd number as the second number gives us a probability of
Multiplying the two, we get
and because both cases have equal probability, we have the probability for the sum to be odd is
Finally, subtracting this from gives us
~ Cheetahboy93
See Also
1987 AJHSME (Problems • Answer Key • Resources) | ||
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