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If in a binomial distribution `n=4, P(X=0)=(16)/(81), t h e n P(X=4)` equals `1/(16)` b. `1/(81)` c. `1/(27)` d. `1/8`

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If in a binomial distribution `n=4, P(X=0)=(16)/(81), t h e n P(X=4)` equals `1/(16)` b. `1/(81)` c. `1/(27)` d. `1/8` A. `(1)/(16)` B. `(1)/(81)` C. `(1)/(27)` D. `(1)/(8)` Select the correct answer from above options

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Correct Answer - B We have, `n=4 and P(X=0)=(16)/(81)` Let p be the probability of success and q that of failure in a trial. Then, `P(X=0)=(16)/(81)` `rArr .^C_(0)q^4=(16)/(81)` ` rArr q^4=((2)/(3))^4rArr q=(2)/(3)rArr p=(1)/(3)` `therefore P(X=4)=.^4C_(4)p^4q^0=p^4=((1)/(3))^4=(1)/(81)`

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