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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

(ii) not more than one will fuse after $150$ days of use.

Answers (1)

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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 $X\leq 1$ ,

$P(X\leq 1)=P(X=0)+P(X=1)$

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

$=(0.95)^{5}+ (0.25)(0.95)^4$

$=(0.95)^{4}(0.95+ 0.25)$

$=(0.95)^{4}\times 1.2$

Posted by

seema garhwal

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