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.

My answer is 1/9
Approach is as follows

1. Total number of ways = 3c2*3c2 = 9
2. No. of pairs of boys such that sum is = 2 O, 1 E = O
3. No. of pairs of Girls = 2 o, 1 e

Case one - For Z to Even
Number of ways - E e, Oo1, Oo2, oO1, oO2 = 5
Probability - 5/9
Case two - For z to be ODD
Number of ways - Eo1, Eo2, eO1, eO2 = 4
Probbility - 4/9
Difference = 1/9

Hope i am using correct approach..

Excellent Solution experts as always. Thank You!!

Anaira Mitch wrote:Excellent Solution experts as always. Thank You!!

I didn't answer, but I have to chime in and speak for the group: we love the appreciation!

But now that I've found the thread, I can't resist a solution of my own.

Once we know that they're distinct, the actual ages of each child aren't relevant, only their parity: whether they're even or odd. With that in mind, we've got two sets:

Boys: {E, E, O}
Girls: {O, E, O}

Choosing four of these, we'll get an odd sum with one odd or three odds, and an even sum with two odds. (Zero odds and four odds aren't possible, of course.)

Since the even sum is easier (only one calculation), let's do that, and find the probability of exactly two odds.

We'll pick two odds in one of two ways:

1) We take the odd boy and exactly one of the odd girls;
2) We don't take the odd boy, but we take both odd girls

Case #1:

Odd Boy * (Exactly One Odd Girl) =

(1 - Both Even Boys) * (1 - Both Odd Girls) =

(1 - 1/3) * (1 - 1/3) = 4/9

Case #2:

Both Even Boys * Both Odd Girls =

(2/3 * 1/2) * (2/3 * 1/2) = 1/9

So our probability of exactly two odds is the sum of these cases, 4/9 + 1/9 = 5/9.

The probability of NOT exactly two odds is 1 - (exactly two odds) = 1 - 5/9 = 4/9.

So the difference between the two is 5/9 - 4/9, or 1/9, and we're done!

ashg84 wrote:My answer is 1/9
Approach is as follows

1. Total number of ways = 3c2*3c2 = 9
2. No. of pairs of boys such that sum is = 2 O, 1 E = O
3. No. of pairs of Girls = 2 o, 1 e

Case one - For Z to Even
Number of ways - E e, Oo1, Oo2, oO1, oO2 = 5
Probability - 5/9
Case two - For z to be ODD
Number of ways - Eo1, Eo2, eO1, eO2 = 4
Probbility - 4/9
Difference = 1/9

Hope i am using correct approach..

Counting the number of cases is often a nice way to solve, so long as you know that the probability of each case is the same. (Here, any group of four children is equally likely, so counting isn't bad.)