A die is tossed thrice. If getting a four is | vedclass

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(D) This is a binomial distribution problem where $n = 3$ (number of trials).Success is defined as getting a $4$ on a die.The probability of success $

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None of these

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(D) This is a binomial distribution problem where $n = 3$ (number of trials).Success is defined as getting a $4$ on a die.The probability of success $p = \frac{1}{6}$.The probability of failure $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.Mean of binomial distribution is given by $\mu = np = 3 	imes \frac{1}{6} = \frac{1}{2}$.Variance of binomial distribution is given by $\sigma^2 = npq = 3 	imes \frac{1}{6} 	imes \frac{5}{6} = \frac{5}{12}$.Thus,the mean is $\frac{1}{2}$ and the variance is $\frac{5}{12}$.Since this pair is not present in options $A, B,$ or $C$,the correct option is $D$.

$A$ die is tossed thrice. If getting a four is considered a success,then the mean and variance of the probability distribution of the number of successes are

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