$A$ carton of $24$ bulbs contains $6$ defective bulbs. One bulb is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced, and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?
(N/A) Total number of bulbs, $n(S) = 24$.
Number of defective bulbs $= 6$.
Number of non-defective (good) bulbs $= 24 - 6 = 18$.
Part $1$: Probability that the bulb is not defective:
$P(\text{not defective}) = \frac{\text{Number of good bulbs}}{\text{Total number of bulbs}} = \frac{18}{24} = \frac{3}{4}$.
Part $2$: If the first bulb selected is defective and not replaced:
Remaining total number of bulbs $= 24 - 1 = 23$.
Remaining number of defective bulbs $= 6 - 1 = 5$.
Probability that the second bulb is defective $= \frac{\text{Remaining defective bulbs}}{\text{Remaining total bulbs}} = \frac{5}{23}$.
Number of defective bulbs $= 6$.
Number of non-defective (good) bulbs $= 24 - 6 = 18$.
Part $1$: Probability that the bulb is not defective:
$P(\text{not defective}) = \frac{\text{Number of good bulbs}}{\text{Total number of bulbs}} = \frac{18}{24} = \frac{3}{4}$.
Part $2$: If the first bulb selected is defective and not replaced:
Remaining total number of bulbs $= 24 - 1 = 23$.
Remaining number of defective bulbs $= 6 - 1 = 5$.
Probability that the second bulb is defective $= \frac{\text{Remaining defective bulbs}}{\text{Remaining total bulbs}} = \frac{5}{23}$.
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