Problem Solving

hgupta0 wrote: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.