Given: A box contains 100 bulbs, 20 of which are defective.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
Ten bulbs are drawn at random for inspection,
Total possible outcomes are 100C10
n (S) = 100C10
(i) Let E be the event that all ten bulbs are defective
n (E) = 20C10
P (E) = n (E) / n (S)
= 20C10 / 100C10
(ii) Let E be the event that all ten good bulbs are selected
n (E) = 80C10
P (E) = n (E) / n (S)
= 80C10 / 100C10
(iii) 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′) = 80C10
P (E′) = n (E′) / n (S)
= 80C10 / 100C10
So, P (E) = 1 – P (E′)
= 1 – 80C10 / 100C10
(iv) Let E be the event that none of the selected bulb is defective
n (E) = 80C10
P (E) = n (E) / n (S)
= 80C10 / 100C10