Question

A poll is given, showing 40% are in favor of a new building project.If 6 people are chosen at random, what is the probability that exactly 3 of them favor the newbuilding project?

Answer 1

SOLUTION

We will use the formula

$\begin{gathered} P_x={^nC_x}p^xq^{n-x} \\ \text{where }P_x=binomial\text{ probability} \\ ^nC_x=\text{ number of combinations }=^6C_3 \\ p=probability\text{ of success on a single try}=\frac{40}{100}=0.4 \\ q=probability\text{ of failure on a single try}=\frac{60}{100}=0.6 \\ x=\text{ }number\text{ of times for a specific }trials\text{ with outcome n = 3} \\ n=\text{ number of trials }=\text{ 6} \end{gathered}$

So we can see that since 40% are in favor of a new building project, then probability of success = 0.4, hence failure becomes 1 - 0.4 = 0.6

Now substituting the figures we have above into the equation, we have

$\begin{gathered} P_x={^nC_x}p^xq^{n-x} \\ P_3={^6C_3}\times(0.4)^3\times(0.6)^{6-3} \\ =\frac{6\times5\times4\times3!}{(6-3)!3!}\times(0.4)^3\times(0.6)^3 \\ =\frac{120}{6}\times0.064\times0.216 \\ 20\times0.064\times0.216 \\ =0.27648 \end{gathered}$

Hence the answer is 0.276 to 3 decimal places

Note that the value of x comes from the statement "exactly 3"