mun23 wrote:
Three boys are ages 4, 6 and 7 respectively. Three girls are ages 5, 8 and 9, respectively. If two of the boys and two of the girls are randomly selected and the sum of the selected children's ages is z, what is the difference between the probability that z is even and the probability that z is odd?
(A) 1/9
(B) 1/6
(C) 2/9
(D) 1/4
(E) 1/2
The sum of the two selected boys can be either 4 + 6 = 10, 4 + 7 = 11 or 6 + 7 = 13. Thus, there is a 1/3 probability that the sum of the ages of the two boys will be even and 2/3 probability that the sum of the ages of the two boys will be odd.
Similarly, the sum of the two selected girls can be either 5 + 8 = 13, 5 + 9 = 14 or 8 + 9 = 17. Thus, there is a 1/3 probability that the sum of the ages of the two girls will be even and 2/3 probability that the sum of the ages of the two girls will be odd.
Now, let’s first find the probability that z is even. Since z is the sum of ages of the selected boys and girls, z can be even if the sum of both the selected boys and selected girls ages are even or if sum of both the selected boys and selected girls ages are odd. The probability that both sums are even is 1/3 x 1/3 = 1/9 and the probability that both sums are odd is 2/3 x 2/3 = 4/9. Thus, there is a 1/9 + 4/9 = 5/9 probability that z is even.
Similarly, let’s find the probability that z is odd. z can only be odd of one of the sums is even and the other is odd. Probability that the boys sum is even and girls sum is odd is 1/3 x 2/3 = 2/9. Probability that the boys sum is odd and the girls sum is even is 2/3 x 1/3 = 2/9. Thus, there is a 2/9 + 2/9 = 4/9 probability that z is odd.
Finally, the difference between the probabilities that z is even and z is odd is 5/9 - 4/9 = 1/9.
Answer: A
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