Question

(i) Among 20 bulbs, 4 bulbs are defective. A bulb is drawn at random from these. What is the probability that the bulb drawn will be defective?

(ii) Let the bulb drawn is neither defective nor it is mixed with the remaining bulbs again. Now a bulb is taken out from these remaining bulbs. What is the probability of the bulb drawn is not defective?

Answer 1

(i) The number of total bulbs = 20

The number of defective bulbs = 4

If a bulb is drawn at random, then the number of favourable outcomes of defective bulbs = 4

Total number of all possible outcomes = 20

∴ The required probability = \(\frac { 4 }{ 20 }\) = \(\frac { 1 }{ 5 }\)

(ii) If the bulb drawn in neither defective nor mixed into remaining bulbs.

Again another bulb is taken out.

∴ The total number of all possible outcomes 

= 20 – 1 = 19

The number of favourable outcomes of the event that the bulb drawn is defective = 4

∴ The probability that the bulb drawn is defective = \(\frac { 4 }{ 19 }\).

∴ The probability that the bulb drawn is not defective 

= 1 – \(\frac { 4 }{ 19 }\) = \(\frac { 15 }{ 19 }\)