Question

A poll is given, showing 30% are in favor of a new building project.

If 5 people are chosen at random, what is the probability that exactly 4 of them favor the new building project?

Answer 1

Answer:

0.02835

Step-by-step explanation:

Let p be the probability who are in favor of the project and q be the probability who are not in favor of the project.

p = 30% = 0.30

q = 1 - p = 1 - 0.30 = 0.70

n = 5, x = 4

Now, by binomial distribution formula:

$P(X=x)=^nC_x \:p^x \:q^{n-x}$

$\implies P(X=4)=^5C_4\: (0.30)^4\: (0.70)^{5-4}$

$\implies P(X=4)=5 (0.30)^4 \:(0.70)^{1}$

$\implies P(X=4)=5 (0.0081) \:(0.70)$

$\implies P(X=4)=0.02835$

The probability that exactly 4 of them favor the new building project is 0.02835.