Hi All,
This question has been posted twice but I'm still not getting the textbook answer to it.

A gamber rolls three fair six-sided dice. What is the probability that two of the dice show the same number, but the third shows a different number?

1. The first die can be ANY number. : I get that so we are comparing the second and third die against this.

2. Second matches first, third does not : 1/6 (there are 6 numbers so the second die has a one in 6 chance of matching the first), third does not X 5/6 (opposite to second, so has a 5 in 6 chance on not matching the first). Is that correct?

The real confusion starts below.

3. Third matches the first, second does not : Third matches first 5/6 (why would this not be 1/6 much like 2 above, is there not a 1/6 chance of the third matching the first?) X second does not1/6 (would this not be 5/6, if the chance the second matches the first is 1/6, why would the chance that it doesn't match also be 1/6)

4. Same as above if the second and third match each other would it not be 1/6 X 1/6.

I hope I'm as clear as mud there. I just need a clear explanation and I'll get it promise.

Thanks in Advance,
Paul

Yes. Chance of first two matching and not the third is 1 x 1/6 x 5/6 = 5/36



Where did you get this solution from? You're right, the 5/6 is the chance the second die doesn't match and the 1/6 is the chance the third die matches. 5/36 again..



Nope. You want the second to be different from the first. 5/6. You want the third to match the second. 1/6. Again, 5/36..

Add these three probabilities together because you want any of the three to occur: 5/36 + 5/36 + 5/36 = 15/36 = 5/12..

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Tim Sanders
Manhattan GMAT Instructor