If a dice is thrown 5 times,then the probabil | vedclass

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(B) This is a binomial distribution problem where $n = 5$ and $k = 3$. The probability of getting a $6$ in a single throw is $p = \frac{1}{6}$. The pr

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$\frac{125}{3888}$

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(B) This is a binomial distribution problem where $n = 5$ and $k = 3$. The probability of getting a $6$ in a single throw is $p = \frac{1}{6}$. The probability of not getting a $6$ is $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$. The probability of getting exactly $k$ successes in $n$ trials is given by the formula $P(X = k) = {}^nC_k \cdot p^k \cdot q^{n-k}$. Substituting the values: $P(X = 3) = {}^5C_3 \cdot (\frac{1}{6})^3 \cdot (\frac{5}{6})^{5-3}$ $P(X = 3) = 10 \cdot \frac{1}{216} \cdot (\frac{5}{6})^2$ $P(X = 3) = 10 \cdot \frac{1}{216} \cdot \frac{25}{36}$ $P(X = 3) = \frac{250}{7776} = \frac{125}{3888}$.

If a dice is thrown $5$ times,then the probability of getting $6$ exactly three times is:

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