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 

            (iii) more than one will fuse after150 days of use.

Answers (1)
S seema garhwal

Let X represent number of bulb 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,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> 1 ,

                        P(X> 1)=1-(P(X=0)+P(X=1))

                                          =1-(^5C_0.(0.95)^{5} . (0.05)^{0}+^5C_1 (0.95)^{4} . (0.05)^{1})

                                          =1-((0.95)^{5}+ (0.25)(0.95)^4)

                                          =1-((0.95)^{4}(0.95+ 0.25))

                                         =1-(0.95)^{4}\times 1.2

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