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Mathematics

(i) A lot of 20 bulbs contains 4 defective bulbs. 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?

Probability

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Answer

No. of defective bulbs = 4.

∴ No. of favourable outcomes = 4

P(getting a defective ball) = No. of favourable outcomesNo. of possible outcomes=420=15\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{20} = \dfrac{1}{5}.

Hence, probability of getting a defective bulb = 15\dfrac{1}{5}.

(ii) Since, one bulb is drawn at random and is not defective and is not replaced.

∴ No. of possible outcomes = 19.

∴ No. of good bulbs present = 15 (16 - 1)

∴ No. of favourable outcomes = 15

P(not getting a defective ball) = No. of favourable outcomesNo. of possible outcomes=1519\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{15}{19}.

Hence, probability of not getting a defective bulb = 1519\dfrac{15}{19}.

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