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Question
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.
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Solution
Let X = number of doublets.
p = probability of getting a doublet when a pair of dice is thrown
∴ p = `6/36 = 1/6` and
q = 1 - p = `1 - 1/6 = 5/6`
Given: n = 4
∴ X ~ B`(4, 1/6)`
The p.m.f. of X is given by
P(X = x) = `"^nC_x p^x q^(n - x)`
i.e. p(x) = `"^nC_x (1/6)^x (5/6)^(4 - x)`, x = 0, 1, 2, 3, 4
∴ P(2 successes) = P(X = 2)
= p(2) = `"^4C_2 (1/6)^2 (5/6)^(4 - 2)`
`= (4!)/(2!*2!) (1/6)^2 (5/6)^2`
`= (4 * 3 * 2!)/(2!.2.1) xx 1/36 xx 25/36`
`= 25/216`
Hence, the probability of two successes is `25/216`.
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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.
∴ P(x ≥ 2) = P(x = 2) + P(x = 3)
= `square` + `square`
= `2/27`
