Two dice are thrown together. The probability | vedclass

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(D) When two dice are thrown,the total number of outcomes is $n(S) = 6 	imes 6 = 36$. Let $E$ be the event that the sum of the numbers is divisible b

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(D) When two dice are thrown,the total number of outcomes is $n(S) = 6 	imes 6 = 36$. Let $E$ be the event that the sum of the numbers is divisible by $2$ or $3$. The possible sums range from $2$ to $12$. Sums divisible by $2$ are: $2, 4, 6, 8, 10, 12$. Sums divisible by $3$ are: $3, 6, 9, 12$. Combining these,the sums divisible by $2$ or $3$ are: $2, 3, 4, 6, 8, 9, 10, 12$. Counting the outcomes for each sum: Sum $2: (1,1) - 1$ outcome Sum $3: (1,2), (2,1) - 2$ outcomes Sum $4: (1,3), (2,2), (3,1) - 3$ outcomes Sum $6: (1,5), (2,4), (3,3), (4,2), (5,1) - 5$ outcomes Sum $8: (2,6), (3,5), (4,4), (5,3), (6,2) - 5$ outcomes Sum $9: (3,6), (4,5), (5,4), (6,3) - 4$ outcomes Sum $10: (4,6), (5,5), (6,4) - 3$ outcomes Sum $12: (6,6) - 1$ outcome Total favorable outcomes $n(E) = 1 + 2 + 3 + 5 + 5 + 4 + 3 + 1 = 24$. Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{24}{36} = \frac{2}{3}$.

Two dice are thrown together. The probability that the sum of the numbers is divisible by $2$ or $3$ is

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