Question

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
II. not more than one
III. more than one
IV. at least one
will fuse after 150 days of use.

Answer 1

Let X represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials.
It is given that, p = 0.05
$\therefore q=1-p=1-0.05=0.95$
X has a binomial distribution with n = 5 and p = 0.05
$\therefore P(X=x)=^nC_xq^{n-x}p^x, \, where \, x=1,2,...n$
= $^5C_X(0.95)^{5-x}.(0.05)^x$

(i) P (none) = P(X = 0)
= $^5C_0(0.95)^5.(0.05)^0$
= $1*(0.95)^5$
=$(0.95)^5$

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(ii) P (not more than one) = P(X ≤ 1)
= P(X=0)+P(X=1)
= $^5C_0(0.95)^5*(0.05)^0+ ^5C_1(0.95)^4*(0.05)^1$
= $1*(0.95)^5*(0.05)^0+^5C_1(0.95)^4*(0.05)^1$
= $1*(0.95)^5+5*(0.95)^4*(0.05)$
= $(0.95)^5+(0.25)(0.95)^4$
= (0.95)^4[0.95+0.25]
=(0.95)^4*1.2

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(iii) P (more than 1) = P(X > 1)
= $1-P(X\leq1)$
= 1-P(not more than 1)
= 1-$(0.95)^4*1.2$

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(iv) P (at least one) = P(X ≥ 1)
= 1-P(X<1)
=1-P(X=0)
=$1- {^5}C_0(0.95)^5*(0.05)^0$
=$1-1*(0.95)^5$
= $1-(0.95)^5$