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If the probability that a fulurorescent light has a useful life of a least 800 hours is 0.9 find the probability that among 20 such lights at least 2 will not have a useful life of at least 800 hours. [ Given `(0.9) = 0.1348 ] ` |
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Answer» Let X = Number of fluorescent light which has useful life. p = Probability that the light will not have useful life. q = Probability that the light has useful life. `:.q=0.9=(9)/(10)` `:.p=1-q=1-(9)/(10)=(1)/(10)` `n=20` (Given) `:.X~B(n,p)` `:.X~B(20,(1)/(10))` The p.m.f. of X is given by `P[X=x]=.^(n)C_(x)p^(x)q^(n-x.)` i.e. ` P(x)=.^(20)C_(x)((1)/(10))^(x)*((9)/(10))^(20-x)` `:.` P (at least 2 lights will not have a useful life ) `=P[X ge 2 ] = 1 - P [ X lt 2 ] ` `=1-[ P (X=0)+P(X=1)]` `=1-[P(0)+P(1)]` `=1-[.^(20)C_(0)((1)/(10))^(0)((9)/(10))^(20-0)+.^(20)C_(1)((1)/(10))^(1)((9)/(10))^(20-1)]` `=1-[1xx1xx((9)/(10))^(20)+20xx(1)/(10)xx((9)/(10))^(19)]` `=1-[(0.9)^(20)+2xx(0.9)^(19)]` `=1-[(0.9+2)*(0.9)^(19)]` `=1-[(2.9)(0.1348)]` `=1-0.3910=0.6090` Hence, the probability that at least 2 lights will not have a useful life of at least 800 hours is 0.6090. |
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