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

Explanation

Let X is the random variable which denotes that a bulb is defective.
Also, n=10,`p=1/50` and q=`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
`thereforep(X=r)=P((0))=""^(10)C_(0)(1/50)^(0)(49/50)^(10-0)=(49/50)^(10)`
(ii) Exactly two bulbs are defectivei.e., r=2
`thereforeP(X=r)=P_((2))=""^(10)C_(2)(1/50)^(2)(49/50)^(8)`
`=(10!)/(8!2!)(1/50)^(2)cdot(49/50)^(8)=45xx(1/50)^(10)xx(49)^(8)`
(iii) More than 8 bulbs work properly i.e., there is less than 2 bulbs which are defective.
So, `rlt2rArrr=0,1`
`thereforeP(X=r)=P(rlt2)=P(0)+P(1)`
`=""^(10)C_(0)(1/50)^(0)(49/50)^(10-0)+""^(10)C_(1)(1/50)^(1)(49/50)^(10-1)`
`=(49/50)^(10)+(10!)/(1!9!)cdot1/50cdot(49/50)^(9)`
`=(49/50)^(10)+1/5cdot(49/50)^(9)=(49/50)^(9)(49/50+1/5)`
`=(49/50)^(9)(59/50)=(59(49)^(9))/((50)^(10))`