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The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. What is the probability that out of 5 such bulbs

Question:

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. What is 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.

Solution:

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

$\therefore \mathrm{P}(\mathrm{X}=x)={ }^{9} \mathrm{C}_{2} q^{x-x} p^{x}$, where $x=1,2, \ldots n$

$={ }^{5} \mathrm{C}_{x}(0.95)^{5-x} \cdot(0.05)^{x}$

(i) P (none) = P(X = 0)

$={ }^{5} \mathrm{C}_{0}(0.95)^{5} \cdot(0.05)^{0}$

$=1 \times(0.95)^{5}$

$=(0.95)^{5}$

(ii) P (not more than one) = P(X ≤ 1)

$=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)$

$={ }^{5} \mathrm{C}_{0}(0.95)^{5} \times(0.05)^{0}+{ }^{5} \mathrm{C}_{1}(0.95)^{4} \times(0.05)^{1}$

$=1 \times(0.95)^{5}+5 \times(0.95)^{4} \times(0.05)$

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

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

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

(iii) P (more than 1) = P(X > 1)

$=1-P(X \leq 1)$

$=1-P($ not more than 1$)$

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

(iv) P (at least one) = P(X ≥ 1)

$=1-\mathrm{P}(\mathrm{X}<1)$

$=1-\mathrm{P}(\mathrm{X}=0)$

$=1-{ }^{5} \mathrm{C}_{0}(0.95)^{5} \times(0.05)^{0}$

$=1-1 \times(0.95)^{5}$

$=1-(0.95)^{5}$

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