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Q. 5: 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 will fuse after 150 days of use.

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Let X represent the number of bulbs that will fuse after 150 days of use. Trials = 5

        P=0.005

        q=1-0.005=1-0.005=0.95

X has a binomial distribution with n = 5.

\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x

                P(X=x)=^5C_x.(0.95)^{5-x} . (0.05)^{x}

             Put X=0 ,

                  P(X=0)=^5C_0.(0.95)^{5} . (0.05)^{0}

                                          =(0.95)^{5}

Posted by

seema garhwal

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