A die is tossed twice. Getting a number great | vedclass

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(B) Let $X$ be the random variable representing the number of successes in $n = 2$ trials.The probability of getting a number greater than $4$ (i.e.,$

Vedclass · Accepted Answer

$\frac{4}{9}$

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(B) Let $X$ be the random variable representing the number of successes in $n = 2$ trials.The probability of getting a number greater than $4$ (i.e.,$5$ or $6$) in a single toss is $p = \frac{2}{6} = \frac{1}{3}$.The probability of failure is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.Since the trials are independent,$X$ follows a binomial distribution $B(n, p)$ with $n = 2$ and $p = \frac{1}{3}$.The variance of a binomial distribution is given by the formula $	ext{Var}(X) = npq$.Substituting the values,we get $	ext{Var}(X) = 2 	imes \frac{1}{3} 	imes \frac{2}{3} = \frac{4}{9}$.

$A$ die is tossed twice. Getting a number greater than $4$ is considered a success. Then the variance of the probability distribution of the number of successes is

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