Three boys are ages 4, 6 and 7 respectively. Three girls are

Question

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

(A) 1/9
(B) 1/6
(C) 2/9
(D) 1/4
(E) 1/2

umeshpatil · Accepted Answer

Age of Boys:4, 6, 7
Sum of ages taken 2 at a time: 10,13,11

Ages of Girls:5, 8, 9
Sum of ages taken 2 at a time: 13,17,14

9 Combinations of sum between sets(10,12,11) & (13,17,14)
=23,27,24- 16,30,17- 24,28,25

Prob(Even)= 5/9
Prob(Odd) =4/9

Answer=5/9 - 4/9 = 1/9

ScottTargetTestPrep · Answer

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

akhandamandala · Answer

I would present another approach.

P(z odd) = P(boys odd)* P(girls even) + P(boys even)* P(girls odd)
 = 2/C2,3 * 1/C2,3 + 1/C2,3 * 2/C2,3
 = 4/9

P(z even) = 1 - P(z odd) = 5/9

|P(z even)-P(z odd)| = 1/9

Konstantin1983 · Answer

I think 13 should be here not 12. And i don't understand second bolded fragment. May be this should be (26,30,27)?

DropBear · Answer

Please feel free to critique me here but this is how I solved.

Boys: 4, 6, 7
Girls: 5, 8, 9
Number of possible combinations of boys or girls = 3!/2! = 3
i.e. there is 3 possible combinations of girls and 3 of boys

Probability that sum of 2 boys ages is even = 1/3 [a]
Probability that sum of 2 boys ages is odd = 2/3 [b]
Probability that sum of 2 girls ages is even = 1/3 [c]
Probability that sum of 2 girls ages is odd = 2/3 [d]

probability that sum of 2 girls and 2 boys is even = [a]*[c] + [b]*[d] = 5/9 [e]
probability that sum of 2 girls and 2 boys is odd = [a]*[b] + [c]*[d] = 4/9 [f]

Therefore the differences in the probabilities is [e] - [f] = 1/9

No idea if this is correct or if it was dumb luck. I am struggling daily with this GMAT journey so would appreciate the feedback.

anairamitch1804 · Answer

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.

We could list all the ways to pick four children—two boys and two girls. We could also take the opposite approach: list all the ways to leave out two children—one boy and one girl. There are fewer scenarios to list with the left-out children, so let's take that approach. The sum of the ages of all six children is (4 + 6 + 7) + (5 + 8 + 9) = 39, an odd number. We can then list all 9 scenarios, subtracting out the ages of the left-out children.

Of the 9 scenarios listed, 5 yield an even z and 4 yield an odd z. Each outcome is equally likely.
The difference between the probability that z is even and the probability that z is odd is therefore 5/9 – 4/9 = 1/9.
The correct answer is A

Table pasted below to represent all 9 scenarios.

Rickooreoisb · Answer

I solved as

In boys 
2 even 1 odd 
In girls 
1 even 2 odd

Thus for boys at even is 2c2 =1 
For girls at odd is 1c1*2c1 = 2 or at even is 2c2=1
So two cases for the even boys 
1*2 giving even = 2
1*1 giving even = 1

Now for odd boys is 2c1(from even set of age)*1c1=2
Again girl is same 
So here 
2*2=4 (odd) and 2*1=2 (even)
2+1+4+2 = 9 cases 
Even is 2+1+2 = 5
Odd is 4

Thus giving 5/9 and 4/9

Is there correct method

Navdeep2000 · Answer

Since there are 3 boys and 3 girls, we know there can be 9 possible combinations by FCP rule.

Let's list them out given that each combination should consist of 2 boys and 2 girls.

(4,6,5,8)= 23  (4,6,5,9)= 24  (4,6,8,9)= 27

(4,7,5,8)= 24  (4,7,5,9)= 25  (4,7,8,9)= 28

(6,7,5,8)= 26  (6,7,5,9)= 27  (6,7,8,9)= 30

So, we have  5 even  and  4 odd  sums i.e. z as total.
Their probability :  Even=(5/9), Odd=(4/9)

Therefore, the difference between them =  (5/9)-(4/9) = (1/9) .

Option A

Three boys are ages 4, 6 and 7 respectively. Three girls are

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Three boys are ages 4, 6 and 7 respectively. Three girls are

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