Binomial Distribution Question

Question

30.4% of US households own one or more cats. You visit five households, selected randomly.

a. What’s the chance that no more than two have cats?

b. What's the chance that exactly two have cats?

c.What's the chance that at least two have cats?

Answer 1

Hi McKenzie!

There are multiple ways to do this. First, I'll tell you how to use your calculator for this. Then, I'll go through the manual steps.

Calculator method

Hit 2ND then VARS to get to the distributions menu. Scroll down til you find binompdf and binomcdf (they're pretty far down). pdf is for a Particular point while cdf is Cumulative. For part a & part c, you'll use cdf. For part b, you'll use pdf. Words like "no more than" or "at least" are the indicators there. "Exactly" will always be pdf.

You'll put in three arguments: first, the sample size (5), then the probability of success (.304), then optionally the # of successes you're looking for. Your calculator for part b should read binompdf(5,0.304,2).

For part A, you'll use binomcdf(). The inputs are the same. The thing to keep in mind is that it will give you a left-tailed probability. It'll always give you the probability of getting at most the # of successes you've input. The phrase "at least" says that it should be a right tailed probability, so you'll have to do 1 minus the probability that at most one house has cats to find the answer for part C.

Manual Method

Part b is the easiest here, so we'll start there.

Let's say there's households A, B, C, D, and E. If we're looking for exactly two, it could be any two. It might be A & B or A & D or B & C. There are a bunch of these combinations. To find out how many, we'll find the binomial coefficient. This might be represented as P(5,2), 5 nCr 2 or ( 52 ). On a TI calculator, you should enter your sample size (in this case 5) then hit MATH and scroll over to PRB. select nCr and then hit 2.

You can do this by hand using the following formula:

n!

r!(n-r)!

where n is sample size and r is your selected number.

So, in this case, 5 nCr 2 is 10.

The probability that two houses have cats is just (.304)²(1-.304)^5-2: ~0.0312. Multiply that by the 20 we found above and you'll find the probability for part b is 31.15%.

For part a, you'll want to repeat the process and add in the probability that one has a cat and the probability that none have cats.

For part c, you could approach it one of two ways: add together the probability that 2, 3, 4, or 5 of the houses have cats, OR add together the probability that 0 or 1 house has cats and just do 1 minus that number.