In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is
$10 to the power of negative 1 end exponent$
$open parentheses 1 half close parentheses to the power of 5$
$open parentheses 9 over 10 close parentheses to the power of 5$
$9 over 10$

Solution

open parentheses 9 over 10 close parentheses to the power of 5

Here n = 5
$straight p space equals space 10 over 100 space equals space 1 over 10 comma space space space straight q space equals space 1 minus straight p space equals space 1 minus 1 over 10 space equals space 9 over 10$
P(none is defective) = P(0) = $straight C presuperscript 5 subscript 0 space open parentheses 9 over 10 close parentheses to the power of 5 space equals space 1 space cross times space open parentheses 9 over 10 close parentheses to the power of 5 space equals space open parentheses 9 over 10 close parentheses to the power of 5$
∴ (C) is correct answer.

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