Probability

Ok, but for the original question, we were able to do 1/2 x 1/2 x 1/2 = 1/8. If I add, because of the "or" clause, I get 1/4, the correct answer.

In the question I posted, if I do 1/2 x 1/2 x 1/2 x 1/2, I get 1/16, but that is not the correct answer. If I add it, because it could be a boy OR a girl, I still only get 1/8, which isn't the correct answer either.

Why is there a difference?

bml1105 wrote:Ok, but for the original question, we were able to do 1/2 x 1/2 x 1/2 = 1/8 which is the correct answer.

In the question I posted, if I do 1/2 x 1/2 x 1/2 x 1/2, I get 1/16, but that is not the correct answer.

Why is there a difference?

The answer to the original problem is 1/4, not 1/8.
In each problem, we have to account for the NUMBER OF WAYS to get a favorable outcome.

Original problem:

The probability of having a girl is identical to the probability of having a boy. In a family with three children, what is the probability that all the children are of the same gender?

a) 1/8.
b) 1/6.
c) 1/3.
d) 1/5.
e) 1/4.

One way to get a favorable outcome is BBB.
P(BBB) = 1/2 * 1/2 * 1/2.
But a favorable outcome can be achieved in the following ways:
BBB
GGG
Since there a total of 2 ways, we multiply by 2:
(1/2 * 1/2 * 1/2) * 2 = 1/4.

The correct answer is E.

Your problem:

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16

One way to get a favorable outcome is BBGG.
P(BBGG) = 1/2 * 1/2 * 1/2 * 1/2.
But a favorable outcome can be achieved in the following ways:
BBGG
BGBG
BGGB
GBBG
GBGB
GGBB
Since there a total of 6 ways, we multiply by 6:
(1/2 * 1/2 * 1/2 * 1/2) * 6 = 3/8.

The correct answer is A.

Hi
I solved it as below:
Prob (g,g,g) + Prob (b,b,b) = 1/2*1/2*1/2 + 1/2*1/2*1/2 = 1/4

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16

The explanation can be understood the following way as well.

2 out of 4 children have to be boy and remaining 2 girls therefore we need to select 2 out of 4 for ensuring them as boys so that remaining two will be girls by default (this also represent different birth orders of two boys and two girls) = 4C2 = 6

Probability of having 2 boys and 2 girls in any order = (1/2)^4 = 1/16

Required probability = 4C2 x (1/16) = 6/16 = 3/8 Option A

P[(G and G and G) or (B and B and B)] = (P[G] * P[G] * P[G]) + (P * P * P) = 0.5^3 + 0.5^3 = 0.25

E