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  • Find the probability of getting 5 exactly twice in 7 throws of a die.

Q. 11  Find the probability of getting 5 exactly twice in 7 throws of a die.

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Let X represent number of times getting 5 in 7 throws of a die.

Probability of getting 5 in single throw of die=P

        P=\frac{1}{6}

        q=1-P=1-\frac{1}{6}=\frac{5}{6}

X has a binomial distribution,n=7

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

                P(X=x)=^7C_x.(\frac{5}{6})^{7-x} . (\frac{1}{6})^{x}

                    P(getting\, \, exactly\, \, twice)=P(X= 2)

                                                                    =^{7}C_2 (\frac{5}{6})^{5}\frac{1}{6}^2

                                                                      =21 (\frac{5}{6})^{5}\frac{1}{36}

                                                                      = (\frac{5}{6})^{5}\frac{7}{12}

Posted by

seema garhwal

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