3% of the electric bulbs manufactured by a company are defective. Using poison distribution (approximately), the probability that a sample of 100 bulbs will contain exactly one defective , is

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.05

.15

e−1

e−2

answer is B.

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Detailed Solution

[p=3100,n=100;∴m=np=3 ]∴ from p(r)= e−m.mrrwe get P(1)=e−m.m11=3e3=antilog[log3−3log10e] =antilog[0.4771−3×0.4343]=antilog (1¯.1742)=0.15

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