$(i)$ $A$ lot of $20$ bulbs contains $4$ 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?
$(ii)$ 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?
(C) $(i)$ Total number of bulbs $= 20$.
Total number of defective bulbs $= 4$.
$P(\text{defective bulb}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{4}{20} = \frac{1}{5}$.
$(ii)$ After drawing one non-defective bulb,the remaining total number of bulbs $= 20 - 1 = 19$.
The number of non-defective bulbs remaining $= (20 - 4) - 1 = 15$.
$P(\text{non-defective bulb}) = \frac{\text{Number of non-defective bulbs}}{\text{Remaining total bulbs}} = \frac{15}{19}$.
Total number of defective bulbs $= 4$.
$P(\text{defective bulb}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{4}{20} = \frac{1}{5}$.
$(ii)$ After drawing one non-defective bulb,the remaining total number of bulbs $= 20 - 1 = 19$.
The number of non-defective bulbs remaining $= (20 - 4) - 1 = 15$.
$P(\text{non-defective bulb}) = \frac{\text{Number of non-defective bulbs}}{\text{Remaining total bulbs}} = \frac{15}{19}$.
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