Discrete probability

Question

7% of all Americans live in poverty. If 39 Americans are randomly selected, find the probability that

a.  Exactly 4 of them live in poverty. 

b.  At most 1 of them live in poverty. 

c.  At least 4 of them live in poverty. 

d.  Between 3 and 7 (including 3 and 7) of them live in poverty. 

Answer 1

Hi Chantal! I'd start by thinking of the poverty status of the this sample as drawn from a binomial distribution with n = 39 and p = .07.

a) Remember that the probability mass function (PMF) of a binomial distribution is (n choose x)* p^x*(1-p)^(n-x). We plug in n = 39, p = .07, and x = 4 to get

(39 choose 4)* (0.07)^4 * (1- 0.07)^(39 - 4)

= 82,251 * .07^4 * (.93)^35

= 0.1558.

b) For this one, all we have to do is calculate P(x = 0) and P(x = 1), and then add them together. Using the same PMF as above:

P(x = 0) = (39 choose 0) * .07^0 * 0.93 ^ 39

= 0.0590

P(x=1) = (39 choose 1) * .07^2 * .93^38

= .0121

Add them together and you get the probability that at most 1 person in the sample lives in poverty: 0.0590 + 0.0121 = 0.0711

c) For this one, we can think of P(at least 4 people live in poverty) as 1 - P(3 or fewer people live in poverty). We already have most of the ingredients to calculate P(3 or fewer live in poverty) - we already have P(x = 0) and P(x = 1) from part b. Now we just need to get P(x = 2) and P(x = 3) using the same PMF like we did above. You'll then get

P(at least four people live in poverty) = 1 - (P(x=0) + P(x=1) + P(x = 2) + P(x = 3)).

d) For this one, you'll have to use the PMF four times: simply add together the following:

P(x=3) + P(x=4) + P(x=5) + P(x=6) + P(x=7)

You'll already have P(x=3) and P(x=4) from part c - you'll just have to use the PMF again to solve for the others. Once you do that, just plug in the numbers and you're golden :)