# Find the Probability of Throwing at Most 2 Sixes in 6 Throws of a Single Die - Mathematics and Statistics

Sum

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

Concept: Bernoulli Trials and Binomial Distribution
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#### APPEARS IN

NCERT Class 12 Maths
Chapter 13 Probability
Q 12 | Page 578