Answer 1

Let X denote the number of sixes.

P(getting a six when a die is thrown) = p = `(1)/(6)`

∴ q = 1 – p = `1 - (1)/(6) = (5)/(6)`

Given, n = 6
∴ X ∼ B`(6, 1/6)`
The p.m.f. of X is given by

P(X = x) = `""^6"C"_x (1/6)^x (5/6)^(6 - x), x` = 0, 1,...,6
P(getting at most 2 sixes)
= P(X ≤ 2)
= P(X = 0 or X = 1 or X = 2)
= P(X = 0) + P(X = 1) + P(X = 2)

= `""^6"C"_0(1/6)^0 (5/6)^6 + ""^6"C"_1(1/6)(5/6)5 + ""^6"C"_2(1/6)^2 (5/4)^4`

= `(5/6)^4 + 6(1/6)(5/6)^5 + (6!)/(2! xx 4!)(1/6)^2 (5/6)^4`

= `(5/6)^4 [(5/6)^2 + (5/6) + (6 xx 5 xx 4!)/(2 xx 1 xx 4) xx (1/6)^2]`

= `(5/6)^4 [25/6^2 + 30/6^ + 15/6^2]`

= `(5/6)^4 (70/36)`

= `(5/6)^4 (35/18)`

= `(5/4)^4 ((5 xx 7)/(6 xx 3))`

= `(7)/(3)(5/6)^5`.