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{\rule{0.5in}{0.1pt}}. (Type a fraction. Simplify your answer.)

Solution

To solve this problem, we’ll use the binomial probability formula. The probability of having exactly \( k \) successes (boys, in this case) in \( n \) trials (births) is given by \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( \binom{n}{k} \) is the binomial coefficient, \( p \) is the probability of success on a single trial, and \( (1-p) \) is the probability of failure. In this problem: - \( n = 8 \) (total number of children) - \( k = 7 \) (number of boys) - \( p = 0.5 \) (probability of having a boy, assuming it’s as likely as a girl) First, calculate the binomial coefficient: \[ \binom{8}{7} = \frac{8!}{7!(8-7)!} = \frac{8}{1} = 8 \] Next, calculate \( p^k \) and \( (1-p)^{n-k} \): \[ p^k = (0.5)^7 = \frac{1}{128} \] \[ (1-p)^{n-k} = (0.5)^1 = \frac{1}{2} \] Now substitute these into the binomial probability formula: \[ P(X = 7) = 8 \times \frac{1}{128} \times \frac{1}{2} = \frac{8}{256} = \frac{1}{32} \] So the probability of having 7 boys and 1 girl in any order is \( \frac{1}{32} \).