Question

assume that boys and girls are equally likely. find the probability that when a couple has three children, there are exactly 3 girls. what is the probability of exactly 3 girls out of three children? (type an integer or a simplified fraction.)

Explanation:

Step1: Determine the probability of a girl

Since boys and girls are equally likely, the probability of having a girl (\(p\)) is \(\frac{1}{2}\), and the probability of having a boy (\(q\)) is also \(\frac{1}{2}\).

Step2: Use the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times q^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successful trials, \(C(n,k)=\frac{n!}{k!(n - k)!}\) is the combination formula. Here, \(n = 3\) (number of children), \(k = 3\) (number of girls), \(p=\frac{1}{2}\), \(q=\frac{1}{2}\).

First, calculate \(C(3,3)\):
\(C(3,3)=\frac{3!}{3!(3 - 3)!}=\frac{3!}{3!0!}=\frac{1}{1}=1\) (since \(0!=1\))

Then, calculate \(p^{k}\) and \(q^{n - k}\):
\(p^{k}=(\frac{1}{2})^{3}=\frac{1}{8}\)
\(q^{n - k}=(\frac{1}{2})^{3 - 3}=(\frac{1}{2})^{0}=1\)

Now, substitute into the binomial formula:
\(P(X = 3)=C(3,3)\times(\frac{1}{2})^{3}\times(\frac{1}{2})^{0}=1\times\frac{1}{8}\times1=\frac{1}{8}\)

Answer:

\(\frac{1}{8}\)