A factory produces bulbs. The probability that any one bulb is defective is

    \[1/50\]

and they are packed in boxes of

    \[10\]

. 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
A factory produces bulbs. The probability that any one bulb is defective is

    \[1/50\]

and they are packed in boxes of

    \[10\]

. 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’s assume X to be the random variable denoting a bulb to be defective.

Here,

    \[n\text{ }=\text{ }10,\text{ }p\text{ }=\text{ }1/50,\text{ }q\text{ }=\text{ }1\text{ }\text{ }1/50\text{ }=\text{ }49/50\]

We know that,

    \[P\left( X\text{ }=\text{ }r \right)\text{ }={{~}^{n}}{{C}_{r}}~{{p}^{r~}}{{q}^{n\text{ }\text{ }r}}\]

(i) None of the bulbs is defective, i.e.,

    \[r=0\]

    \[P\left( x\text{ }=\text{ }0 \right)\text{ }={{~}^{10}}{{C}_{0}}~{{\left( 1/50 \right)}^{0}}{{\left( 49/50 \right)}^{10\text{ }\text{ }0}}~=\text{ }{{\left( 49/50 \right)}^{10}}\]

(ii) Exactly two bulbs are defective

So,

    \[P\left( x\text{ }=\text{ }2 \right)\text{ }={{~}^{10}}{{C}_{2}}~{{\left( 1/50 \right)}^{2}}{{\left( 49/50 \right)}^{10\text{ }\text{ }2}}\]

=

    \[{{45.49}^{8}}/{{50}^{10}}~=\text{ }45\text{ }\times \text{ }{{\left( 1/50 \right)}^{10}}~\times \text{ }{{49}^{8}}\]

(iii) More than

    \[8\]

bulbs work properly

We can say that less than

    \[2\]

bulbs are defective

    \[P\left( x\text{ }<\text{ }2 \right)\text{ }=\text{ }P\left( x\text{ }=\text{ }0 \right)\text{ }+\text{ }P\left( x\text{ }=\text{ }1 \right)\]