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Question

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,

will fuse after 150 days of use.

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Solution

Let P be probability of success and q be the probability of failure

P = 0.05, q = 0.95

n = 5

(i)n(x=0)=nC1p0qn0=nC0q5=(0.95)5(ii)p(x1)=p(x=0)+p(x=1)=5C0p0q5+5C1p1q4=(0.95)5+5(0.05)(0.95)4


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