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Question
A die is thrown 6 times. If ‘getting an odd number’ is a success, find the probability of at least 5 successes.
Solution
Let X = number of successes, i.e. number of odd numbers.
p = probability of getting an odd number in a single throw of a die
∴ p = `3/6 = 1/2` and q = `1 - "p" = 1 - 1/2 = 1/2`
Given: n = 6
∴ X ∼ B`(6, 1/2)`
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 (1/2)^x (1/2)^(6 - x)`
` = "^6C_x (1/2)^6,` x = 0, 1, 2, ...,6
P(at least 5 successes) = P[X ≥ 5]
= p(5) + p(6)
`= ""^6C_5 (1/2)^6 + "^6C_6 (1/2)^6`
`= (""^6C_5 + "^6C_6) (1/2)^6`
`= (6 + 1) 1/64 = 7/64`
Hence, the probability of at least 5 successes is `7/64`.
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