Question

As voters exit the polls, you ask a representative random sample of voters if they voted for a proposition. If the true percentage of voters who vote for the proposition is 63%, what is the probability that, in your sample, exactly 5 do not voted for the proposition before 2 voted for the proposition? The probability is

Answer 1

Answer:

The probability is 0.057797

Step-by-step explanation:

Consider the provided information.

It is given that true percentage of voters who vote for the proposition is 63%,

Let p is probability of success.

According to the binomial distribution:

$P(x;p,n)=^nC_x(p)^x(1-p)^{(n-x)}$

Substitute n=7, p=0.63 and x=2 in the above formula.

$P(x;p,n)=^7C_2(0.63)^2(1-0.63)^{(7-2)}$

$P(x;p,n)=\frac{7!}{2!5!}(0.3969)(0.37)^{5}\\P(x;p,n)=21(0.3969)(0.37)^{5}\\P(x;p,n)=0.0577974947199\\P(x;p,n)\approx0.057797$

Hence, the probability is 0.057797