(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 bulb is defective? (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
(i) Total number of bulbs = 20. Number of defective bulbs = 4. Probability of drawing a defective bulb is calculated as (Number of favorable outcomes) / (Total number of outcomes). Therefore, Probability = 4/20 = 1/5.
(ii) Since the bulb drawn in (i) is not defective and is not replaced, the total number of bulbs decreases. Original number of good bulbs = 20 - 4 = 16. After drawing one good bulb, remaining good bulbs = 15. Total remaining bulbs = 19. Probability that the bulb is not defective = 15/19.
Explanation
The solution follows the standard probability formula P(E) = Number of favorable outcomes / Total number of outcomes. For part (i), the favorable outcome is selecting one of the 4 defective bulbs from the lot of 20. For part (ii), the problem involves 'without replacement', similar to Question 15 in the context. Since one non-defective bulb is removed, the count of good bulbs reduces from 16 to 15, and the total count reduces from 20 to 19.