Question

The Wilson family had 9 children. Assuming that the probability of a child being a girl is 0.5, find the probability that the Wilson family had: at least 2 girls? at most 3 girls? Round your answers to 4 decimal places.

Answer 1

X : Number of girl children that Wilson Family had

Number of children Wilson family had : n = 9

Probability of a child being a girl :p = 0.5

X follows a Binomial distribution with n=9 and p=0.5 (q=1-p=1-0.5=0.5)

Probability that Wilson Family had 'r' girl children = P(X=r)

Probability that the Wilson family had at least 2 girls = P(X2)

P(X2) = 1 - P(X<2) = 1-P(X1)

P(X1) = P(X=0)+P(X=1)

P(X1) = P(X=0)+P(X=1) = 0.001953125+0.017578125 = 0.01953125

P(X2) = 1 - P(X<2) = 1-P(X1) = 1-0.01953125 =0.98046875

Probability that the Wilson family had at least 2 girls = P(X2) = 0.98046875

Probability that the Wilson family had at least 2 girls = 0.9805

Probability that the Wilson family had at most 3 girls = P(X 3)

P(X 3) = P(X=0)+P(X=1)+P(X=2)+P(X=3)

P(X 3) = P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.001953125+0.017578125+0.0703125+0.1640625=0.25390625

Probability that the Wilson family had at most 3 girls = P(X 3) = 0.25390625

Probability that the Wilson family had at most 3 girls = 0.2539