(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?
The total number of bulbs in the lot is given as 20. Out of these, 4 bulbs are defective. To find the probability that a randomly drawn bulb is defective, we divide the number of defective bulbs by the total number of bulbs. Therefore, the required probability is 4/20, which simplifies to 1/5.
In the second scenario, the bulb drawn in part (i) was not defective and was not replaced. Originally, there were 16 good bulbs (20 total - 4 defective). After removing one good bulb, the remaining total becomes 19, and the remaining good bulbs become 15. The probability of drawing a non-defective bulb now is the ratio of remaining good bulbs to the remaining total, which is 15/19.
Explanation
This question tests the basic understanding of probability in sample spaces with and without replacement. Part (i) uses the standard formula P(E) = Number of favorable outcomes / Total outcomes. Part (ii) requires adjusting the sample space and favorable outcomes based on the condition that one good bulb has already been removed, similar to the logic in Question 15 and 16 of the provided context.