Two unbiased dice are thrown once. Write all | vedclass

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(B) When two unbiased dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$. The sample space $S$ is given by:$S = \{(1, 1), (1,

Vedclass · Accepted Answer

$\frac{5}{12}$

Vedclass · Answer

(B) When two unbiased dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$. The sample space $S$ is given by:$S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$Let $E$ be the event that the sum of the numbers on the two dice is a prime number. The possible prime sums are $2, 3, 5, 7,$ and $11$.The outcomes favorable to event $E$ are:Sum $= 2: (1, 1)$Sum $= 3: (1, 2), (2, 1)$Sum $= 5: (1, 4), (2, 3), (3, 2), (4, 1)$Sum $= 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$Sum $= 11: (5, 6), (6, 5)$Total number of favorable outcomes $= 1 + 2 + 4 + 6 + 2 = 15$.Therefore,the probability $P(E) = \frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{15}{36} = \frac{5}{12}$.

Two unbiased dice are thrown once. Write all the possible outcomes of this experiment. What is the probability that the sum of numbers on two dice is a prime number?

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