In a box containing 100 bulbs,10 are defectiv | vedclass

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(B) Total number of bulbs = $100$. Number of defective bulbs = $10$. Number of non-defective bulbs = $100 - 10 = 90$. The probability of picking a non

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$\left(\frac{9}{10}\right)^5$

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(B) Total number of bulbs = $100$. Number of defective bulbs = $10$. Number of non-defective bulbs = $100 - 10 = 90$. The probability of picking a non-defective bulb in a single draw is $p = \frac{90}{100} = \frac{9}{10}$. Since we are selecting $5$ bulbs,and assuming the selection is done with replacement (or the sample size is small relative to the population),the probability that none of the $5$ bulbs are defective is given by the binomial probability $P(X=0) = \left(\frac{9}{10}\right)^5$. Thus,the correct option is $B$.

In a box containing $100$ bulbs,$10$ are defective. The probability that out of a sample of $5$ bulbs,none is defective is . . . . . . .

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