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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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