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
not more than one

(iii) more than one

(iv) at least one will fuse after 150 days of use.

.: We have, $p = 0.05 = \cfrac{5}{{100}} = \cfrac{1}{{20}}$

and $q = 1 - \cfrac{1}{{20}} = \cfrac{{19}}{{20}}$

$X$ has a binomial distribution with $n = 5,p = \cfrac{1}{{20}},q = \cfrac{{19}}{{20}}$

$\therefore$ $P(X = r){ = ^n}{C_r}{q^{n - r}}{p^r}$

(i) $P(X = 0){ = ^5}{C_0}{q^5}{p^0} = 1 \times {q^5} \times 1 = {q^5} = {\left( {\cfrac{{19}}{{20}}} \right)^5} = {(0.95)^5}$

(ii) $P(X = 0) + P(X = 1){ = ^n}{C_0}{q^n}{p^0}{ + ^n}{C_1}{q^{n - 1}}p$

$= 1 \times {q^5} \times 1{ + ^5}{C_1}{q^4}p = {\left( {\cfrac{{19}}{{20}}} \right)^5} + 5{\left( {\cfrac{{19}}{{20}}} \right)^4}\left( {\cfrac{1}{{20}}} \right)$

$= {\left( {\cfrac{{19}}{{20}}} \right)^4}\left( {\cfrac{{19}}{{20}} + \cfrac{5}{{20}}} \right) = {\left( {\cfrac{{19}}{{20}}} \right)^4}\left( {\cfrac{{24}}{{20}}} \right) = \cfrac{6}{5}{\left( {\cfrac{{19}}{{20}}} \right)^4} = (1.2){(0.95)^4}$

(iii) $P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$= 1 - (P(X = 0) + P(X = 1)) = 1 - \cfrac{6}{5}{\left( {\cfrac{{19}}{{20}}} \right)^4} = 1 - (1.2){(0.95)^4}$

[Using part (ii)]
(iv) $P(X = 1) + P(X = 2) + \ldots \ldots \ldots \ldots \ldots + P(X = 5)$

$= 1 - P(X = 0) = 1 - {\left( {\cfrac{{19}}{{20}}} \right)^2} = 1 - {(0.95)^5}$

[Using part (i) ]