GMAT Practice Question: 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?
- \frac{3}{8}
- \frac{1}{4}
- \frac{3}{16}
- \frac{1}{8}
- \frac{1}{16}
Topics: probability, p&c multisource
Solution
Step 1
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(\text{boy}) = \frac{1}{2} P(\text{girl}) = \frac{1}{2}
Step 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
Step 3
We count how many sequences have exactly two girls by selecting two positions out of four.
\frac{4!}{2!(4-2)!} = \frac{4\times3\times\cancel{2\times1}}{\cancel{2\times1}\times2\times1} = \frac{4\times3}{2\times1} = 6
Step 4
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}
Step 5
We divide the number of favorable sequences by the total sequences and then simplify the fraction.
\frac{6}{16} = \frac{\cancel{2}\times3}{\cancel{2}\times8} = \frac{3}{8}
Answer
\frac{3}{8}
