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# Help plz

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Each of five, standard, six-sided dice is rolled once. Two of the dice come up the same, but the other three are all different from those two and different from each other. The pair is set aside, and the other three dice are re-rolled. The dice are said to show a "full house" if three of the dice show the same value and the other two show the same value (and potentially, but not necessarily, all five dice show the same value). What is the probability that after the second set of rolls, the dice show a full house?

Nov 11, 2018

#2
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You can choose the two paired dice in (5C2) ways, the value of the pair in 6 ways and the non-paired dice in 5*4*3 ways, giving us this:
5C2*6*5*4*3 / 6^5 =25/54
Now we roll the three odd dice to obtain a three of a kind of any value.
First die - any number 6/6
2nd die - 1/6 prob. of matching the first
3rd die - 1/6 prob. of matching the first two:
P =6/6 * 1/6 * 1/6 =1/36
Now will multiply these together to get the final probability: 25/54 * 1/36 =25/1944=0.01286

Nov 11, 2018
#3
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But what if the first die out of the 3 is the same as the first 2 that were already rollled, and the last 2 are different. That would work too, right?

Guest Nov 11, 2018
#4
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Also, it is given that the pair and 3 different already happened, you don't need to calculate that probability.

Guest Nov 11, 2018
#5
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Those answers are incorrect. Here is the correct one:

We divide this into cases, the 3 are the same, one is the same as the original and the last 2 are the same, and they are all the same.

Case 1: 5 ways

Case 2: 3*5=15 ways

Case 3: 1 way

Total: 21 ways

total without restrictions:6^3

So probability is 21/216=7/72.

Nov 11, 2018