Jason and Maria are playing a board game ....

I'm not that good at probability, but I'll take a run at this one...

P(Two 5's) = Two dice show "5" and one other shows any number but 5.

So, we can choose one of the three dice to show 5 and on that die it can show 5 possible things.  So we have C(3,1)*5 = 15. And since there are 216 possible outcomes, we have

15/216 = 5/72.

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P(At Least Two 5s).....This means we want to find:

P(Two 5s) + P(Three 5s)

There's only one way for all three dice to show three 5s. So we have:

15/216 + 1/216 = 16/216 = 2/27

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P(One 5 or Two 5s) = P(One 5) + P(Two 5s) - P(One 5 and Two 5s)

The last term = 0 (Since we can't roll one 5 and two 5s on the same roll)

For the probability of rolling one 5, note that the other two dice can show any numbers but 5, and there are 5 ways * 5 ways = 25 ways to do this. And we can choose one of the three die to show the 5. So we have:

25 * C(3,1) = 75   ...And again, there are 216 possible outcomes....So

P(One 5 or Two 5s) = 75/216 + 15/216 = 90/216 = 5/12

I think Melody and I only differed on the second answer.......Any arbitrators out there?? I'll abide by any decision the jury reaches........

P(exactly 2 fives)=(1/6 * 1/6 * 5/6) * 3C2 = 5/216 * 3 = 15/216 = 5/72

P(at least 2 5s) = 1-P(no5s) = 1- (5/6)^3 = 1-(125/216) = 91/216

The above answer is wrong.  Mine is a lest one 5.  CPhill's answer is correct.

P(one5 or two fives)=P(1five)+P(2fives)

= 3C1 * (1/6*5/6*5/6) + 15/216 = 3*25/216 + 15/216 = 90/216 = 5/12

Thanks Chris.