Probability2

[Abdulla]

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


OA is [spoiler]3*5/36=15/36[/spoiler]

[logitech]

lets solve it for 1 number and multiply it by 6 ( for all possible numbers )

Same dice number = 1

1 1 (5) = 5 ways

1 (5) 1 = 5 ways

(5) 1 1 = 5 ways

So 15 ways for one number

15x6 ways for all six numbers

All the possible ways = 6x6x6

P = What we want / what could happen

= 15x6/6x6x6

= 15/36

I just came back from Vegas, I am good at this!

[Abdulla]

Good job..

If Vegas will help me to pass the F* GMAT I'll go tomorrow..

[hgupta0]

Can someone elaborate on the answer please.

Based on the answer above, it seems that the probability of getting two number correct and a third one wrong is greater than 1/3, which seems to high to me.

The way i did this was:

(1/6)*(1/6)*(5/6) and a combination of this - which we can multiply by 3.

Why is the answer different?

Any help would be appreciated.

- H

[Ian Stewart]

Can someone elaborate on the answer please.

The way i did this was:

(1/6)*(1/6)*(5/6) and a combination of this - which we can multiply by 3.

- H

That's almost right- if you wanted to know the probability of getting, say, two 4's and one number which was different from 4, then your calculation is perfect. You need to multiply by six, though, because there are six numbers on a die, and any one of these could be our repeated number.

At first glance, 15/36 seemed high to me as well, but you can see that it must be correct in another way. When you roll three dice, three things can happen:

all are different: Probability = (6/6)*(5/6)*(4/6) = 20/36

all are same: Probability = (6/6)*(1/6)*(1/6) = 1/36 (the first roll doesn't matter, as long as the next two rolls match)

two are same, one different: Probability = 1 - the two probabilities above = 1 -(20/36 + 1/36) = 15/36

logitech's solution is certainly faster, and is what I'd do on the test, but I think there's value in seeing a variety of approaches to counting/probability questions - makes it more likely you'll be able to see at least one path to a solution on test day.