Question

andrew is one of nine children. what is the probability that exactly five of the children are males?

Explanation:

Step1: Identify the distribution

This is a binomial probability problem. We assume that the probability of a child being male (p) is $0.5$ (and female is also $0.5$), the number of trials (n) is $9$ (total children), and we want the probability of $k = 5$ successes (males). The binomial probability formula is $P(X=k)=\binom{n}{k}p^{k}(1 - p)^{n - k}$.

Step2: Calculate the combination

First, calculate the combination $\binom{n}{k}=\binom{9}{5}$. The formula for combinations is $\binom{n}{k}=\frac{n!}{k!(n - k)!}$. So, $\binom{9}{5}=\frac{9!}{5!(9 - 5)!}=\frac{9!}{5!4!}=\frac{9\times8\times7\times6\times5!}{5!\times4\times3\times2\times1}=\frac{9\times8\times7\times6}{4\times3\times2\times1}=126$.

Step3: Calculate the probability

Now, $p = 0.5$, $1-p = 0.5$, $n = 9$, $k = 5$. Plug into the binomial formula: $P(X = 5)=\binom{9}{5}(0.5)^{5}(0.5)^{9 - 5}=126\times(0.5)^{5}\times(0.5)^{4}=126\times(0.5)^{9}$. Since $(0.5)^{9}=\frac{1}{512}$, then $126\times\frac{1}{512}=\frac{126}{512}=\frac{63}{256}\approx0.2461$.

Answer:

$\frac{63}{256}$ (or approximately $0.2461$)