Question

a couple plans to have three children. what is the probability that
a) they have all boys?
b) they have at least one girl?

Explanation:

Step1: Determine total number of outcomes

Each child can be either a boy (B) or a girl (G). So for 3 children, the total number of possible combinations is $2\times2\times2 = 2^3=8$ (by the multiplication principle).

Step2: Calculate probability of all - boys

The event of having all boys is just 1 outcome (BBB). So the probability $P(\text{all boys})=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{1}{8}$.

Step3: Calculate probability of at - least one girl

The complement of the event “at least one girl” is the event “all boys”. Let $A$ be the event of having at least one girl. Then $P(A)=1 - P(\text{all boys})$. Since $P(\text{all boys})=\frac{1}{8}$, then $P(A)=1-\frac{1}{8}=\frac{7}{8}$.

Answer:

a) $\frac{1}{8}$
b) $\frac{7}{8}$