Question

Find the probability of throwing at most 2 sixes in 6 throws of a single die.

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 - 0) + ""^6"C"_1(1/6)^1(5/6)^(6 - 1) + ""^6"C"_2(1/6)^2 (5/6)^(6 - 2)`

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

= `(5/6)^4 [(5/6)^2 + 5/6 + 5/12]`

= `(5/6)^4 [25/36 + 30/36 + 15/36]`

P(X ≤ 2) = `(5/6)^4[(25 + 30 +15)/36]`

P(X ≤ 2) = `(5/6)^4 (70/36)`

P(X ≤ 2) = `(5/6)^4 (35/18)`

P(X ≤ 2) = `(5/6)^4 ((5 xx 7)/(6 xx 3))`

P(X ≤ 2) = `(7/3)(5/6)^5`