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प्रश्न
A die is thrown 6 times. If ‘getting an odd number’ is a success, find the probability of at most 5 successes.
उत्तर
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 most 5 successes) = P[X ≤ 5]
= 1 - P [X > 5]
= 1 - p(6) = `1 - "^6C_6 (1/2)^6`
`= 1 - 1 xx 1/64 = 63/64`
Hence, the probability of at most 5 successes is `63/64`.
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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)
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Probability of getting at least two success means x ≥ 2.
∴ P(x ≥ 2) = P(x = 2) + P(x = 3)
= `square` + `square`
= `2/27`
