Question

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

Answer 1

The repeated tossing of the die are Bernoulli trials. Let X represent the number of times of getting sixes in 6 throws of the die.

Probability of getting six in a single throw of die, `p = 1/6`

`therefore q = 1 - p = 1 - 1/6 = 5/6`

Clearly, X has a binomial distribution with n = 6

The p.m.f. of X is given by

P(X = x) = `"^nC_x  p^x q^(n - x)`

i.e. p(x) = `"^6C_x (1/6)^x (5/6)^(6 - x)`, x = 0, 1, 2, ....,6

P(at most 2 sixes) = P[X ≤ 2]

= p(0) + p(1) + p(2)

`= ""^6C_0 (1/6)^0 (5/6)^(6 - 0) + ""^6C_1 (1/6)^1 (5/6)^(6 - 1) + "^6C_2 (1/6)^2 (5/6)^(6 - 2)`

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

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

`= (5/6)^6 + (5/6)^5 + 15 xx 1/36 xx (5/6)^4`

`= [(5/6)^2 + (5/6) + 15/36](5/6)^4`

`= (25/36 + 5/6 + 15/36).(5/6)^4`

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

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

`= 7/3 xx 10/12 xx (5/6)^4`

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

Hence, the probability of throwing at most 2 sixes

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