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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 - Mathematics and Statistics

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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APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 8 Probability Distributions
Exercise 8.3 | Q 1.05 | Page 150
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