1. ## Probability Question

Hello!

I was hoping such a forum existed, and with the magic of Google, I found it I'm very interested in understanding this as much as possible, but Im not very mathematically inclined.

Attached is the table from my textbook. The question states as follows:

"Suppose that two people were randomly selected from the referenced population. Given that one of them had been divorced, what is the probability that the other one had been divorced?"

The answer given in the book is .2246 but every way Ive tried to solve, I do not come to this answer. I feel like I maybe am wrong in what the events would be listed. Logically I want to say Event A would be person 1 is divorced and Event B is person 2 is divorced.

What am I missing here?

2. ## Re: Probability Question

There are two fairly simple approaches I can see that one could take to solve this problem. They both give the same answer of course. But let's use the first approach I'm thinking of: Direct probability calculations.

The question we want to answer takes the form: P(Both people chosen are divorced | At least one of them is divorced). Using the definition of conditional probability we can rewrite this as:

P(Both people chosen are divorced AND at least one of them is divorced) / P(At least one of them is divorced)

The numerator clearly reduces to P(Both people chosen are divorced). We can then look at what the probability of both being divorced is

P(Both people chosen are divorced)
= (# of ways we can choose two divorced people) / (# of ways we can choose two people from this population)
= Choose(612, 2) / Choose(1669, 2)

Where Choose(n, k) = n!/(k!(n-k)!)

The denominator is P(At least one of them is divorced) - we'll take a less direct approach to calculating this by recalling that P(A) = 1 - P(A') where A' is the complement of A.

P(At least one of them is divorced) = 1 - P(It is not true that at least one of them is divorced) = 1 - P(Neither are divorced)

If we're looking to calculate P(Neither are divorced) we can use a similar calculation as before:

P(Neither are divorced)
= (# of ways to choose two people that haven't been divorced) / (# of ways to choose two people from this population)
= Choose(1057, 2) / Choose(1669, 2)

So the probability of interest is:
[Choose(612,2) / Choose(1669,2)] / [1 - Choose(1057,2)/Choose(1669,2)]

3. ## The Following User Says Thank You to Dason For This Useful Post:

Owl Salesman (02-16-2014)

4. ## Re: Probability Question

Thanks for the info.

The text states to break it down in the form of conditional probability rule so P(B|A) = P(A and B)/P(A)

I've done other problems of similar format but for whatever reason arriving at an answer at this one keeps stumping me.

5. ## Re: Probability Question

Originally Posted by Owl Salesman
The text states to break it down in the form of conditional probability rule so P(B|A) = P(A and B)/P(A)
Indeed. That's what we did.

The other approach I had in mind is a little easier to get one's head around but isn't as generalizable.

6. ## The Following User Says Thank You to Dason For This Useful Post:

Owl Salesman (02-16-2014)

7. ## Re: Probability Question

Ah ok I see it there. Can you explain this portion a bit more?

P(Both people chosen are divorced)
= (# of ways we can choose two divorced people) / (# of ways we can choose two people from this population)
= Choose(612, 2) / Choose(1669, 2)

Where Choose(n, k) = n!/(k!(n-k)!)

I think my problem isnt so much the equations themselves as it is conceptualizing where the data itself comes from. I tended to struggle with word problems in Algebra, but in this case I find statistics more interesting because of its real world application...so I want to have a strong understanding.

 Tweet

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts