The number of trials conducted in a binomial | vedclass

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(D) For a binomial distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1 - p$.Given $n = 6$ and $\mu - \sigma^2 = \fra

Vedclass · Accepted Answer

$\frac{154}{4^6}$

Vedclass · Answer

(D) For a binomial distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1 - p$.Given $n = 6$ and $\mu - \sigma^2 = \frac{27}{8}$.Substituting the formulas: $np - npq = \frac{27}{8} \implies np(1 - q) = \frac{27}{8}$.Since $1 - q = p$,we have $np^2 = \frac{27}{8}$.Substituting $n = 6$: $6p^2 = \frac{27}{8} \implies p^2 = \frac{27}{48} = \frac{9}{16}$.Thus,$p = \frac{3}{4}$ and $q = 1 - \frac{3}{4} = \frac{1}{4}$.The probability of getting $X$ successes is given by $P(X = k) = \binom{n}{k} p^k q^{n-k}$.We need to find $P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)$.$P(X = 0) = \binom{6}{0} (\frac{3}{4})^0 (\frac{1}{4})^6 = 1 	imes 1 	imes \frac{1}{4^6} = \frac{1}{4^6}$.$P(X = 1) = \binom{6}{1} (\frac{3}{4})^1 (\frac{1}{4})^5 = 6 	imes \frac{3}{4^6} = \frac{18}{4^6}$.$P(X = 2) = \binom{6}{2} (\frac{3}{4})^2 (\frac{1}{4})^4 = 15 	imes \frac{9}{4^6} = \frac{135}{4^6}$.Summing these: $P(X \le 2) = \frac{1 + 18 + 135}{4^6} = \frac{154}{4^6}$.

The number of trials conducted in a binomial distribution is $n = 6$. If the difference between the mean and variance of this variate is $\frac{27}{8}$,then the probability of getting at most $2$ successes is:

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