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(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?

Answer :

(i) Numbers of defective bulbs = 4

The total numbers of bulbs = 20

\( P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)

\(\therefore \) Probability of getting a defective bulb = P (defective bulb) \( = \ \frac{4}{20} \ = \ \frac{1}{5} \ = \ 0.2 \)

(ii) Since 1 non-defective bulb is drawn, then the total numbers of bulbs left are 19

So, the total numbers of events (or outcomes) = 19

Numbers of defective bulbs = 19 - 4 = 15

So, the probability that the bulb is not defective \( = \ \frac{15}{19} \ = \ 0.789 \)

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