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Q. 14 In a box containing $100$ bulbs, $10$ are defective. The probability that out of a sample of $5$ bulbs, none is defective is

          (A)   $10^{-1}$

              (B)   $\left ( \frac{1}{5} \right )^{5}$

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

              (D)   $\frac{9}{10}$

Answers (1)

best_answer

Let X represent a number of defective bulbs out of 5 bulbs.

Probability of getting a defective bulb =P

$P=\frac{10}{100}=\frac{1}{10}$

$q=1-P=1-\frac{1}{10}=\frac{9}{10}$

X has a binomial distribution,n=5

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$ $(\frac{9}{10})^{5-x} . (\frac{1}{10})^{x}$

$P(non\,of \,bulb\,is\, defective \,)=P(X=0)$

$=^{5}C_0 (\frac{9}{10})^{5}\frac{1}{10}^0$

$=1.\frac{9}{10}^5$

$=(\frac{9}{10})^5$

The correct answer is C.

Posted by

seema garhwal

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