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Sum

The probability that a bulb produced by a factory will fuse after 200 days of use is 0.2. Let X denote the number of bulbs (out of 5) that fuse after 200 days of use. Find the probability of X ≤ 1

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#### Solution

Let X denote the number of bulbs that will fuse after 200 days.

P(bulb will fuse after 200 days) = p = 0.2

∴ q = 1 – p = 1 – 0.2 = 0.8

Given, n = 5

∴ X ∼ B(5, 0.2)

The p.m.f. of X is given by

P(X = x) = `""^5"C"_x (0.2)^x (0.8)^(5 - x), x` = 0.1,...,5

P(X ≤ 1) = P(X = 0 or X = 1)

= P(X = 0) + P(X = 1)

= `(0.8)^5 + ""^5"C"_1(0.2)(0.8)^4` ...[From (i)]

= (0.8)^{4} [0.8 + 5 x 0.2]

= (1.8) (0.8)^{4}

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