Question

the accompanying table describes results from groups of 8 births from 8 different sets of parents. the random variable x represents the number of girls among 8 children. complete parts (a) through (d) below. click the icon to view the table. a. find the probability of getting exactly 6 girls in 8 births. (type an integer or a decimal. do not round.)

Explanation:

Step1: Use 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, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 8$ (number of births), $k = 6$ (number of girls), and assuming the probability of having a girl $p=0.5$ (equally likely to have a boy or a girl).
First, calculate the combination $C(8,6)=\frac{8!}{6!(8 - 6)!}=\frac{8!}{6!2!}=\frac{8\times7}{2\times1}=28$.

Step2: Calculate the binomial probability

$P(X = 6)=C(8,6)\times(0.5)^{6}\times(1 - 0.5)^{8 - 6}$. Substitute $C(8,6)=28$, $p = 0.5$, and $n - k=2$ into the formula.
$P(X = 6)=28\times(0.5)^{6}\times(0.5)^{2}=28\times(0.5)^{8}$.
Since $(0.5)^{8}=\frac{1}{256}$, then $P(X = 6)=28\times\frac{1}{256}=\frac{28}{256}=0.109375$.

Answer:

$0.109375$