BTGmoderatorLU wrote:Source: Manhattan Prep
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 OA is A.
Since the ages are integers, their sum would be either even or odd.
We see that there are three boys and three girls, thus, there are 3*3 = 9 possible sums; out of which few are even and few are odd.
Let's first find out the sum = Odd.
Sum would be odd if an even adds to odd OR and odd adds to even.
1. Say, we pick one boy with even age, thus, we must pick a girl with odd age. There are two boys with even age (4 and 6) and there are two girls with odd age (5 and 9).
> Number of ways to select one out of two boys = 2C1 = 2
> Number of ways to select one out of two girls = 2C1 = 2
Total number of ways = 2*2 = 4
2. Say, we pick one boy with odd age, thus, we must pick a girl with even age. There is only one boy with even age (7) and there is only one girl with odd age (8).
> Number of ways to select one out of one boy = 1
> Number of ways to select one out of one girl = 1
Total number of ways = 1*1 = 2
Total number of ways to get odd sum = 4 + 1 = 5
=> The probability that z is odd = 5/9
=> The probability that z is even = 1 - 5/9 = 4/9
Difference = |4/9 - 5/9| = 1/9
The correct answer:
A
Hope this helps!
-Jay
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