Question: In a family with 8 children, excluding multiple births, what is the probability of having 7 boys…

In a family with 8 children, excluding multiple births, what is the probability of having 7 boys and 1 girl, in any order? Assume that a boy is as likely as a girl at each birth. … The probability of having 7 boys and 1 girl is \boxed{}. (Type a fraction. Simplify your answer.)

Solution

To find the probability of having 7 boys and 1 girl in any order when 8 children are born, we use the binomial probability formula. The probability \( p \) of having a boy or a girl is \( 0.5 \). The total number of trials \( n \) is 8, and we want 7 successes (boys), so \( k = 7 \). The binomial probability formula is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] First, calculate \(\binom{n}{k}\), the number of ways to choose 7 boys out of 8 children: \[ \binom{8}{7} = \frac{8!}{7!(8-7)!} = \frac{8 \times 7!}{7! \times 1!} = 8 \] Now substitute the values into the formula: \[ P(X = 7) = \binom{8}{7} (0.5)^7 (0.5)^1 \] \[ P(X = 7) = 8 \times (0.5)^7 \times (0.5) \] \[ P(X = 7) = 8 \times (0.5)^8 \] \[ P(X = 7) = 8 \times \frac{1}{256} \] \[ P(X = 7) = \frac{8}{256} \] Simplify the fraction: \[ P(X = 7) = \frac{1}{32} \] The probability of having 7 boys and 1 girl in a family of 8 children is \(\frac{1}{32}\).