A die is rolled three times. The probability | vedclass

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(C) When a die is rolled three times,the total number of outcomes is $6 	imes 6 	imes 6 = 216$. We are looking for the sum $S$ such that $S$ is a pr

Vedclass · Accepted Answer

$\frac{5}{36}$

Vedclass · Answer

(C) When a die is rolled three times,the total number of outcomes is $6 	imes 6 	imes 6 = 216$. We are looking for the sum $S$ such that $S$ is a prime number of the form $4n+1$. The possible sums range from $3$ to $18$. The prime numbers in this range are $3, 5, 7, 11, 13, 17$. Among these,the primes of the form $4n+1$ are $5, 13, 17$. Number of ways to get sum $5$: $(1,1,3)$ in $3$ permutations,$(1,2,2)$ in $3$ permutations. Total = $6$ ways. Number of ways to get sum $13$: $(1,6,6)$ in $3$ permutations,$(2,5,6)$ in $6$ permutations,$(3,4,6)$ in $6$ permutations,$(3,5,5)$ in $3$ permutations,$(4,4,5)$ in $3$ permutations. Total = $21$ ways. Number of ways to get sum $17$: $(5,6,6)$ in $3$ permutations. Total = $3$ ways. Total favorable outcomes = $6 + 21 + 3 = 30$. Required probability = $\frac{30}{216} = \frac{5}{36}$.

$A$ die is rolled three times. The probability of getting their sum equal to a prime number of the form $4n+1$ is

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