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# Probability 12

Two dice are rolled n times, find the probability that each of the six results (1,1),(2,2),(3,3)...,(6,6) appears at least once.

ANS: summation( k=0→ 6) (-1)^k ( 6Ck) (1- k/36)^n

### 2 Answers

- 自由自在Lv 71 decade agoFavorite Answer
Please refer to Inclusion-ExclusionPrinciple in 參考資料

Let us use A(k) to represent the case when (k,k) does not appear in the n times rolls.

Use | mean OR

Counts of all possibilities = 36^n

Count of A(1)|A(2)|A(3)|A(4)|A(5)|A(6), i.e. at least one of the 6 is missing X=

Sum(one missing)

- Sum(2 missing at same time)

+ Sum(3 missing at the same time)

- Sum(4 missing at same time)

+ Sum(5 missing at the same time)

-Sum(all 6 missing together)

Sum of 1missing = (6C1)(35)^n

6C1: choose 1 from 6 possibilities (1,1),(2,2)...(6,6)

35 because 36 ways minus the one that is missing.

Similarly for 2, 3, 4, etc.

X=(6C1)35^n - (6C2)34^n + (6C3)33^n - (6C4)32^n + (6C5)31^n - (6C6)30^n

Count of all present = All combinations - Count for at least one missing

= 36^n - (6C1)35^n + (6C2)34^n - (6C3)33^n + (6C4)32^n - (6C5)31^n + (6C6)30^n

=Sum(k=0 to 6)(-1)^k(6Ck)(36-k)^n

Required probability = [Sum(k=0 to 6)(-1)^k(6Ck)(36-k)^n]/36^n

=Sum(k=0 to 6)(-1)^k(6Ck)(1-k/36)^n

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- 六呎將軍Lv 71 decade ago
Obviously, the result is valid for n >= 6.

Analysis is as follows:

First of all, in each throw, the probability of appearance of each of (1, 1), (2, 2), ..., (6, 6) is 1/36.

So, out of 6 of n results, they should be (1, 1), (2, 2), ..., (6, 6) with each appearing once.

Therefore, we don't care about the remaining (n - 6) results.

So for n throws, including the consideration of order of appearance, no. of possibilities to satisfy the result is nP6.

And each possibility has its probability = (1/36)6.

Finally, the required answer is nP6/366.

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