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The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs(i)       none(ii)      not more than one(iii)     more than one(iv)     at least onewill fuse after 150 days of use.

Text Solution

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Let X = number of fuse bulbs. p = probability of a bulb produced by a factory will fuse after 150 days of use.
`∴ p = 0.05 and q = 1 – p = 1 – 0.05 = 0.95`
Given: `n = 5
∴ X ~ B(5, 0.05)`
(i) P(none of a bulb produced by a factory will fuse after 150 days of use) = P[X = 0]
`P(none)=P(X=0) = 5^C^0(0.95)^5 ⋅(0.05)^0`
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