Question

A couple has 4 children. Find each probability of a. All girls b. Exactly two girls and two boys c. At least one child who is a girl d. At least one child of each gender

Answer 1

If a couple has 4 children then possible outcomes:

(GGGG, GGGB, GGBG, GBGG, BGGG, BBGG, GGBB, GBBG, BGGB, BGBG, GBGB, BBBG, BBGB, BGBB, GBBB, BBB)

Total number of possible outcomes = 16

a). All girls:

The probability of all are girls would be:

$P(GGGG)=(\frac{1}{2})^4=\frac{1}{16}$

b). Exactly two girls and two boys:

$P(2G \& 2B)=\frac{6}{16}=\frac{3}{8}$

c). At least one girl:

$P(1G\& 3B)=1-P(BBBB)=1-\frac{1}{16}=\frac{15}{16}$

d). At least one child of each gender:

$P(at least 1Gor1B)=\frac{14}{16}=\frac{7}{8}$