Question

A state senator believes that 25% of all senators on the finance committee will strongly support the tax proposal she wishes to advance. suppose that this belief is correct and that 5 senators are approached at random. what is the probability that 3 out of 5 senators will strongly support the proposal?

Answer 1

$P(X=x)= nC_xp^x(1-p)^{n-x}$Number of Senators be n=5
Probability that the Senators will strongly support the tax proposal be $p=\frac{25}{100}=0.25$
Let X be the number of Senators will strongly support the tax proposal.
Using Binomial Distribution,
$P(X=x)=nC_xP^x(1-p)^{n-x}$
Now Plug in the values of n and p,
$P(X=x)=nC_xp^x(1-)^{n-x} [\tex] To find the probability that 3 out of the 5 Senators will support the tax proposal is [tex] P(X= 3)=5C_3(0.25)^3(1-0.25)^{5-3}\\ =0.08789$