ras-j wrote:Q. Three boys ages 4, 6 and 7 respectively. Three girls 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 age is z, what is the difference between the probability that z is even and the probability that z id odd?
A. 1/9
B. 1/6
C. 2/9
D. 1/4
E. 1/2
Total ways to choose the 4 children:
Number of pairs of boys that can be formed from 3 options = 3C2 = 3.
Number of pairs of girls that can be formed from 3 options = 3C2 = 3.
To combine these options, we multiply:
3*3 = 9.
Ways to get an ODD sum:
For the sum of 4 integers to be odd, either ONE of the integers or THREE of the integers must be odd.
Case 1: ONE integer is odd
(4,6) from the boys and (5,8) from the girls, yielding 4+6+5+8 = 23.
(4,6) from the boys and (8,9) from the girls, yielding 4+6+8+9 = 27.
Case 2: THREE integers are odd
(4,7) from the boys and (5,9) from the girls, yielding 4+7+5+9 = 25.
(6,7) from the boys and (5,9) from the girls, yielding 6+7+5+9 = 27.
Total ways to get an ODD sum = 4.
Thus, P(odd sum) = (total ways to get an odd sum)/(total ways to choose the 4 children) = 4/9.
Since P(odd sum) = 4/9, P(even sum) = 1 - 4/9 = 5/9.
Thus:
P(even sum) - P(odd sum) = 5/9 - 4/9 = 1/9.
.
The correct answer is
A.
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