statistics problem

Question

According to a recent survey, 4 % of the population owns a cat. 67 Percent of cat owners are married, while 56 % of non-cat owners are married. If you randomly select a person, what is the probability that she/he is married? 

Answer 1

The first thing to note is that you're selecting randomly, so (at least in Intro Stats) it's safe to assume that (Probability of...) and (Percent of the Population such that...) are the same number after converting between percent and decimal. This won't come up in the calculations, so it's often not mentioned by tutors/instructions/textbooks in the solution.

For example, P(Cat) = 4% = 0.04 from the given information. We'll use this a bit later.

In "67 Percent of cat owners are married," the of is a sign that this is a conditional probability. It may help to read this as "if someone owns a cat, there is a 67% = 0.67 probability that she/he is married. Thus:

P(Married | Cat) = 0.67

Similarly, P(Married | No Cat) = 56% = 0.56.

It's common to get stuck here, but if you look at what you know, you might realize that you can use the Multiplication Rule for Probability:

P(Married AND Cat) = P(Cat) x P(Married | Cat) = .04 x .67 = .0268

But what can you do with P(Married | No Cat)? If you knew P(No Cat), then you could do something similar to above. Fortunately, using Complementary events:

P(No Cat) = 1 - P(Cat) = 0.96

Thus:

P(Married AND No Cat) = P(No Cat) x P (Married | No Cat) = 0.5376

Finally, you can compute P(Married) by noting that

[Married] = [Married and Cat] OR [Married and No Cat]

and using the Addition Rule for Disjoint Events:

P(Married) = .0268 + .5376 = .5644

There are more clever solutions that require less work, but I prefer to show students this one since it's a good review of many different concepts and formulas.