binomial probabilities

Question

A local county has an unemployment rate of 7.9%. A random sample of 17 employable people are picked at random from the county and are asked if they are employed. The distribution is a binomial. Round answers to 4 decimal places.

1. Find the probability that exactly 3 in the sample are unemployed.

Find the probability that there are fewer than 3 in the sample are unemployed.

Find the probability that there are more than 1 in the sample are unemployed.

Find the probability that there are at most 3 in the sample are unemployed.

Answer 1

Hi Alma,

Did you see my answer to your baseball problem? If so, this problem uses the same formula:

P(X=a)=C(n,x)p^xq^n-x

Breaking this down for the first part of your question:

P=probability we are trying to find

X=a=3

n=17

p=probability of unemployment=0.079

q=probability of employment=1-p=0.921

Substituting:

P(X=3)=C(17,3)*0.079³*0.921^17-3

C(17,3)=17!/(3!*14!)

!=Factorial=17*16*15*14….

You an input the quotient above into a TI-83 or greater calculator, but I can help if your instructor requires calculation by hand.

Anyhow:

C(17,3)=17!/(3!*14!)=680

Substituting and simplifying:

P(X=3)=680*0.079³*0.921¹⁴

P(X=3)=0.1059

That answers the first part of your question. For the second part, fewer than 3 implies 2,1, or 0 people unemployed. So calculate:

P(X=2)=C(17,2)*0.079²*0.921¹⁵

P(X=1)=C(17,1)*0.079¹*0.921¹⁶

P(X=0)=C(17,0)*0.079⁰*0.921¹⁷

I will leave those calculations to you, but the final step of the second part is to add those three probabilities together:

P(X<3)=P(X=2)+P(X=1)+P(X=0)

P(X<3)=0.2470+0.3599+0.2468

P(X<3)=0.8537

Now, for the third part, 3 at most unemployed. This is:

P(X<=3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)

From the previous parts, P(X<3)=0.8537, P(X=3)=0.1059

Thus:

P(X<=3)=0.8537+0.1059

P(X<=3)=0.9596

I hope this helps. Good luck.