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17.(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 ?

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Total number of bulbs =  20

Hence, total possible outcomes = 20

Number of defective bulbs = 4

Hence, the number of favourable outcomes = 4

\therefore P(getting\ a\ defective\ bulb)= \frac{favourable\ outcomes}{total\ outcomes} = \frac{4}{20}

= \frac{1}{5}

\therefore P(getting\ a\ non\ defective\ bulb)= 1 - \frac{1}{5} = \frac{4}{5}

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HARSH KANKARIA

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