Question

(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this blb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and not replaced Now, bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer 1

(i) Total number of bulbs = 20

Total number of defective bulbs = 4

P (getting a defective bulb) = $\frac{numberoffavourableoutcomes}{totalnumberofoutcomes}$ = $\frac{4}{20}$ = $\frac{1}{5}$

(ii) Remaining total number of bulbs = 19

Remaining total number of non-defective bulbs = 16 − 1 = 15

P (getting a not defective bulb)=$\frac{15}{19}$