Given is a box contains 100 bulbs, 20 of which are defective.

By using the formula of probability, we get,

P (E) = favourable outcomes / total possible outcomes

Ten bulbs are drawn at random for inspection,

Total possible outcomes are ${}^{100}C_{10}$

$n (S) = {}^{100}C_{10}$

(i) Let E be the event that at least one bulb is defective

E= {1,2,3,4,5,6,7,8,9,10} where 1,2,3,4,5,6,7,8,9,10 are the number of defective bulbs

Let E′ be the event that none of the bulb is defective

$n (E^{′}) = {}^{80}C_{10}$

P (E′) = n (E′) / n (S)

$= {}^{80}C_{10} / {}^{100}C_{10}$

So, P (E) = 1 – P (E′)

$= 1 – {}^{80}C_{10} / {}^{100}C_{10}$

(ii) Let E be the event that none of the selected bulb is defective

$n (E) = {}^{80}C_{10}$

P (E) = n (E) / n (S)

$= {}^{80}C_{10} / {}^{100}C_{10}$