Basic Probability Question

[rwinston]

Hi guys

This one is kind of embarrassing, but its driving me crazy! I am working through some examples in a prob. book, to try and refresh my rusty stats and prob knowledge. There is a question that goes: "There are 8 olive, 4 black, and six brown socks in a drawer. Two are selected at random. What is (a) the probability that the two socks are the same color? (b) If they are the same color, what is the prob. that they are both olive?"

I can figure out part (a), which is:

$\frac{\dbinom{4}{2}+\dbinom{8}{2}+\dbinom{6}{2}}{\dbinom{18}{2}}$

But I can't figure out P(olive|same color). Can anyone help?

Thanks
Rory

Oh BTW, this isn't a homework question - I can see the answer from the back of the book is 4/7 - I am just curious to see how the author got it! Thanks.

 

[matt grime]

How many pairs of socks of the same colour are there? (Hint: you've alread worked that out) How many pairs of olive socks are there? (Hint: you've already worked that out as well).


You can read off the answer from what you wrote above.

 

[rwinston]

Hi Matt

thanks for the reply. I think I am being dense here, but I still can't see the solution - there are (8/2) + (4/2) + (6/2) = 9 unique pairs, and 4 of those pairs are olive. So I would have thought it would be more like (4/9)?

Rory

 

[matt grime]

9 pairs? By your logic, if I have 3 socks, then there are 1.5 pairs of socks. There aren't there are 3. 3 choose 2. Not 3/2. If they're labelled a,b,c then the pairs are (a,b) (a,c) and (b,c). You used the binomial coefficients in the first post, so why have you stopped using them now?

 

[robert Ihnot]

You can use conditional probability: P(B|A)={P(A) intersection P(B)}/P(A).

The probability of B given A is equal to the probability of A intersection B divided by the probability of A.

 

[rwinston]

Thanks for the help - it dawned on me eventually

$\frac{\dbinom{8}{2}}{\dbinom{4}{2}+\dbinom{8}{2}+\dbinom{6}{2}}$

 

[robert Ihnot]

What has happened is that the original sample space was 18x17/2 = 153 (pairs). The new sample space is only those cases where the pairs match, which is 49 pairs. Then we want to find the cases where the pairs are olive, which is 28 pairs, giving us the correct figure of 4/7. Completely logical problem.