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Question
Let X ~ B(10, 0.2). Find P(X ≥ 1).
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Solution
X ~ B(10, 0.2)
∴ n = 10, p = 0.2
∴ q = 1 − p = 1 − 0.2 = 0.8
The p.m.f. of X is given by
P(X = x) = nCx px qn−x
∴ P(X = x) = 10Cx (0.2)x (0.8)10−x ...[x = 0, 1, 2, 3, ...., 10]
P(X ≥ 1) = 1 − P(X < 1)
= 1 − P(X = 0)
= 1 − 10C0 (0.2)0 (0.8)10−0
= 1 − 1 × 1 × (0.8)10
= 1 − (0.8)10
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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`
