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प्रश्न
Let X ~ B(10, 0.2). Find P(X ≥ 1).
उत्तर
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) = `"^nC_x p^x q^(n - x)`
∴ P(X = x) = `"^10C_x (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 - `"^10C_0 (0.2)^0 (0.8)^(10 - 0)`
`= 1 - 1 xx 1 xx (0.8)^10`
`= 1 - (0.8)^10`
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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`
