Question

it is estimated that 4000 of the 10000 viting residents of a town are against a new sales task. if 15 eligible voters are selected at random and asked their opinion, what is the probability that at most 7 favor for the new tax?

Answer 1

4000 out of 10000 voting residents are against a new sales tax.

hence 10000 - 4000= 6000 resident favour the new tax.

the probability that randomly selected voter favor the new tax is

$p= \frac{6000}{10000}=0.6$

15 voters are selected at random

n=15

1-p=1-0.6=0.4

Here random variable x is a binomial random variable with parameter n=15 and p=0.6

The probability distribution of the binomial random variable is

$P(X=x) = \binom{n}{x}p^x(1-p)^{n-x}$

Now find the probability that at most 7 out of 15 randomly selected voters favor the new tax.

$P(X=x) = \binom{15}{x}0.6^x(0.4)^{n-x} \\ P(X≤7) = \sum^7_{i=0}P(X=i) \\ = P(X=0) + P(X=1)+...+P(X=7) \\ = \binom{15}{0}0.6^0(0.4)^{15-0} + \binom{15}{1}0.6^1(0.4)^{15-1}+...+\binom{15}{7}0.6^7(0.4)^{15-7} \\ = 0.2131$

The probability that at most 7 out of 15 randomly selected voters favor the new tax is 0.2131.