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Question
A die is thrown 6 times. If ‘getting an odd number’ is a success, find the probability of 5 successes.
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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(5 successes) = P[X = 5]
`= p(5) = "^6C_5 (1/2)^6`
`= "^6C_1 xx 1/64` ...`[because ""^nC_x = "^nC_(n - x)]`
`= 6/64`
= `3/32`
Hence, the probability of 5 successes is `3/32`
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Solution:
A pair of dice is thrown 3 times.
∴ n = 3
Let x = number of success (doublets)
p = probability of success (doublets)
∴ p = `square`, q = `square`
∴ x ∼ B (n, p)
P(x) = nCxpx qn–x
Probability of getting at least two success means x ≥ 2.
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= `square` + `square`
= `2/27`
If X is a binomial variable with range {0, 1, 2, 3, 4} and P(X = 3) = 3P(X = 4) then the parameter ‘p’ of the binomial distribution is
