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