Question

For next year, 59% of veterans will buy a single-family home. If we pick a random sample of 5 veterans, find the probability that exactly 2 veterans will buy a single-family home next year?

P(X

=

binompdf(,,)

		Percent rounded to two decimal places
P(X	≈	%

Answer 1

Given that next year, 59% of the veterans will buy a single-family home.

Hence, probability that a veteran will buy a single family home is p = 59% = 0.59

Now if we pick a random sample of 5 veterans, we need to find the probability that exactly 2 veterans will buy a single-family home next year.

Let, X = Number of veterans who will buy a single-family home.

$\Rightarrow X\sim Binomial(n=5,p=0.59)$

Now before we go on to solve the problem let us know a bit about Binomial Distribution.

Binomial Distribution

A discrete random variable X is said to have a binomial distribution if its PMF (Probability Mass Function) is given by,

n fx (2) = P(X = 1) = C)p'(1 – p)"-- ,I=0,1,2,. 0 otherwise ?

where 0<p<1.

Notation : X~Binomial n. p)

Coming back to our problem

X = Number of veterans who will buy a single-family home.

$\Rightarrow X\sim Binomial(n=5,p=0.59)$

$f_{X}(x)=P(X=x)=\left\{\begin{matrix} \binom{5}{x}0.59^{x}(1-0.59)^{5-x} &,x=0,1,2,3,4,5 \\ 0 &,otherwise \end{matrix}\right.$

Now here we need to find the probability that exactly 2 veterans will buy a single-family home next year out of the randomly selected 5 veterans.

$P(X=2)$

$=\binom{5}{2}0.59^{2}(1-0.59)^{5-2}$

$=\frac{5!}{2!*(5-2)!}*0.59^{2}(1-0.59)^{3}$

$=\frac{5!}{2!*3!}*0.59^{2}(1-0.59)^{3}$

$=\frac{5!}{2!*3!}*0.59^{2}*0.41^{3}$

$\Rightarrow P(X=2)=0.2399$

$\mathbf{\Rightarrow P(X=2)=23.99\%}$