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