A factory produces bulbs. The probability that any one bulb is defective is $\frac{1}{{50}}$ and they are packed in 10 boxes. From a single box, find the probability that

(i) none of the bulbs is defective.

(ii) exactly two bulbs are defective.

(iii) more than 8 bulbs work properly.

Let X is the random variable which denotes that a bulb is defective.

Also, $n = 10,p = \frac{1}{{50}}$

and $q = \frac{{49}}{{50}}$ and $P(X = r){ = ^n}{C_r}{p^r}{q^{n - r}}$

(i) None of the bulbs is defective i.e., $r = 0$

$\therefore$ $p(X = r) = {P_{(0)}}{ = ^{10}}{C_0}{\left( {\frac{1}{{50}}} \right)^0}{\left( {\frac{{49}}{{50}}} \right)^{10 - 0}} = {\left( {\frac{{49}}{{50}}} \right)^{10}}$

(ii) Exactly two bulbs are defective i.e., $r = 2$

$P(X = r) = {P_{(2)}}{ = ^{10}}{C_2}{\left( {\frac{1}{{50}}} \right)^2}{\left( {\frac{{49}}{{50}}} \right)^8}$

$= \frac{{10!}}{{8!2!}}{\left( {\frac{1}{{50}}} \right)^2} \cdot {\left( {\frac{{49}}{{50}}} \right)^8} = 45 \times {\left( {\frac{1}{{50}}} \right)^{10}} \times {(49)^8}$

(iii) More than 8 bulbs work properly i.e., there is less than 2 bulbs which are defective.

So, $r < 2 \Rightarrow r = 0,1$
$\therefore$ $P(X = r) = P(r < 2) = P(0) + P(1)$

${ = ^{10}}{C_0}{\left( {\frac{1}{{50}}} \right)^0}{\left( {\frac{{49}}{{50}}} \right)^{10 - 0}}{ + ^{10}}{C_1}{\left( {\frac{1}{{50}}} \right)^1}{\left( {\frac{{49}}{{50}}} \right)^{10 - 1}}$

$= {\left( {\frac{{49}}{{50}}} \right)^{10}} + \frac{{10!}}{{1!9!}} \cdot \frac{1}{{50}} \cdot {\left( {\frac{{49}}{{50}}} \right)^9}$

$= {\left( {\frac{{49}}{{50}}} \right)^{10}} + \frac{1}{5} \cdot {\left( {\frac{{49}}{{50}}} \right)^9} = {\left( {\frac{{49}}{{50}}} \right)^9}\left( {\frac{{49}}{{50}} + \frac{1}{5}} \right)$

$= {\left( {\frac{{49}}{{50}}} \right)^9}\left( {\frac{{59}}{{50}}} \right) = \frac{{59{{(49)}^9}}}{{{{(50)}^{10}}}}$