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Sun 21 Oct, 2012 07:38 am

The probability that n dice are rolled is 1/2^n. Let SN denote the sum of the numbers shown on N dice. Find the probability of the following events: a) SN = 4, knowing that N is even; b) N = 2, given that SN = 3; c) N = 2, given that SN = 4 and the rst die showed the number 1; d) the largest number shown on any die is 4, knowing that N = 3.

Can somebody please help me with this?

For point a, we have two cases: N=2 and N=4. When N=2 we can get (1,3) (2,2) and (3,1) and N=4 (1,1,1,1) for favourable cases.

However, I don't know where the 1/2^n part comes in? The total number of cases should be 6^N, right? So 6^2 + 6^4 = 1332. Then, the number of favorable cases/total number of cases should give 6/1332? I think I'm missing something important here, please help me out! Thanks

@retek,

The probability that N is even is 1/3 (1/4 + 1/16 + 1/64 + ...).

The probability that N is 2 is 1/4.

The probability that N is 4 is 1/16.

The probability of rolling 4 with 2 dice is 3/6^2.

The probability of rolling 4 with 4 dice is 1/6^4.

Given that N is even, the probability of rolling 4 is:

[(1/4 * 3/6^2) + (1/16 * 1/6^4)] / (1/3)

which is :

[(P(N=2) * P(rolling 4 with 2 dice)) + (P(N=4) * P(rolling 4 with 4 dice))] / P(N is even)