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 - child event

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

Step2: Use the binomial probability formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of “successes” (in this context, a “success” is having a girl), $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 3$ (number of children), $k = 0$ (number of girls), and $p=\frac{1}{2}$. First, calculate the combination $C(3,0)=\frac{3!}{0!(3 - 0)!}=\frac{3!}{3!}=1$. Then, $p^{k}=(\frac{1}{2})^{0}=1$ and $(1 - p)^{n - k}=(1-\frac{1}{2})^{3-0}=(\frac{1}{2})^{3}=\frac{1}{8}$.

Step3: Calculate the probability

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

Answer:

$\frac{1}{8}$