Question

assume that boys and girls are equally likely. find the probability that when a couple has three children, there are exactly 0 girls. what is the probability of exactly 0 girls out of three children? (type an integer or a simplified fraction.)

Explanation:

Step1: Determine probability of a single - birth outcome

The probability of having a boy (or a girl) in a single birth is $p=\frac{1}{2}$, and the probability of having a girl is also $q = \frac{1}{2}$.

Step2: Use the binomial probability formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{n - k}\times q^{k}$, where $n$ is the number of trials, $k$ is the number of “successes” (in this case, number of girls), $p$ is the probability of non - success, and $q$ is the probability of success. Here, $n = 3$, $k = 0$, $p=\frac{1}{2}$, and $q=\frac{1}{2}$. The binomial coefficient $C(n,k)=\frac{n!}{k!(n - k)!}$, so $C(3,0)=\frac{3!}{0!(3 - 0)!}=\frac{3!}{3!}=1$.

Step3: Calculate the probability

$P(X = 0)=C(3,0)\times(\frac{1}{2})^{3-0}\times(\frac{1}{2})^{0}=1\times(\frac{1}{2})^{3}\times1=\frac{1}{8}$.

Answer:

$\frac{1}{8}$