The probability that a bulb produced by a factory will fuse after 150 days if used is 0.50. What is the probability that out of 5 such bulbs none will fuse after 150 days of use?

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1−(19/20)5

(19/20)5

(3/4)5

90 (1/4)5

answer is B.

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Detailed Solution

Here p = 19/20, q = 1/20, n = 5, r = 5. The required probability is=5C5192051200=19205

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