A couple decides to have 4 children. If they succeed in having 4 children and each child is...

Question

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?

GMAT Panda · Accepted Answer

3/8 We note that each child’s gender is independent of the others and each child has an equal chance to be a boy or a girl. P(boy) = 1/2 P(girl) = 1/2 We determine the total number of possible gender sequences in four births by multiplying the number of outcomes for each birth. 2 2 2 2 = 16 We count how many sequences have exactly two girls by selecting two positions out of four. 4!/2!(4-2)! = 43212121 = 43/21 = 6 We find the probability of any specific sequence of four births as the product of individual probabilities since births are independent. (1/2) (1/2) (1/2) (1/2) = 1/16 We divide the number of favorable sequences by the total sequences and then simplify the fraction. 6/16 = 2328 = 3/8

Explanation	Calculations	Help
$We note that each child’s gender is independent of the others and each child has an equal chance to be a boy or a girl.$	$P (boy) = \frac{1}{2} P (girl) = \frac{1}{2}$
$We determine the total number of possible gender sequences in four births by multiplying the number of outcomes for each birth.$	$2 \times 2 \times 2 \times 2 = 16$
$We count how many sequences have exactly two girls by selecting two positions out of four.$	$\frac{4 !}{2 ! ( 4 - 2 )!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} = \frac{4 \times 3}{2 \times 1} = 6$
$We find the probability of any specific sequence of four births as the product of individual probabilities since births are independent.$	$(\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{16}$
$We divide the number of favorable sequences by the total sequences and then simplify the fraction.$	$\frac{6}{16} = \frac{2 \times 3}{2 \times 8} = \frac{3}{8}$

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