Question

In the US, 50.8% of the people are female. Suppose that 10 people in the US were chosen at random. Find the probability that 6 of those people are female.

Answer 1

Given:

The female percentage is 50.8%

That is,

$P=0.508$

The number of people was chosen at random, n = 10.

To find: The probability that 6 of those people are female.

Explanation:

Let us take,

$x=6$

Using the binomial distribution,

$P(X=x)=^nC_xP^x(1-P)^{n-x}$

On substitution we get,

$\begin{gathered} P(X=6)=^{10}C_6(0.508)^6(1-0.508)^{10-6} \\ =\frac{10!}{(10-6)!6!}(0.508)^6(0.492)^4 \\ =\frac{6!\times7\times8\times9\times10}{1\times2\times3\times4\times6!}(0.508)^6(0.492)^4 \\ =7\times3\times10\times(0.508)^6(0.492)^4 \\ =0.2115 \end{gathered}$

Hence, the probability that 6 of those people are female is 0.2115.

Final answer: 0.2115.