Please show work A die is rolled 6 times Find the probabilit

Please show work

A die is rolled 6 times. Find the probability of rolling Exactly 4 twos Exactly 5 threes At most 3 fours

Solution

Solution:

For a fair six-sided die, the probability of a 2 in any given throw is 1/6, and so the probability of not obtaining a 2 in a throw is 1 - 1/6 = 5/6.

The probability that the first four throws are 2\'s, and the remaining 2 throws are not, is:

1/6×1/6×1/6×1/6×5/6×5/6

=25/46656

But this is not exactly what we want. The 2\'s should be \"allowed\" to occur anywhere in the 6 throws. That is to say, any 4 of the 6 different throws can be 2\'s, and the remaining 2 throws should be anything other than 2. Of course, for any fixed arrangement of four 2\'s and 2 non-2\'s, the probability is the same as above. But how many such arrangements are possible? As any such arrangement is determined by the placement of the four 2\'s, the number of arrangements is merely the number of ways in which we can select four places out of twenty, that is (⁶C₄).

Since the probability of each arrangement is 25/46656, the probability that one of them occurs is (⁶C₄)(25/46656)

Similarly the second part.