Two dice are thrown and the sum of the number | vedclass

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Question

(B) When two dice are thrown,the total number of outcomes is $6 	imes 6 = 36$. Let $S$ be the sum of the numbers on the two dice. The possible values

Vedclass · Accepted Answer

$\frac{13}{36}$

Vedclass · Answer

(B) When two dice are thrown,the total number of outcomes is $6 	imes 6 = 36$. Let $S$ be the sum of the numbers on the two dice. The possible values for $S$ are $\{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$. Event $A$ is the event that the sum is a prime number: $A = \{2, 3, 5, 7, 11\}$. Event $B$ is the event that the sum is greater than $8$: $B = \{9, 10, 11, 12\}$. We need to find $P(A \cap \overline{B})$,which is the probability of the event $A$ occurring and $B$ not occurring. This is equivalent to the event that the sum is a prime number $AND$ the sum is less than or equal to $8$. $A \cap \overline{B} = \{2, 3, 5, 7\}$. Now,we count the number of outcomes for each sum: Sum $= 2: (1,1) ightarrow 1$ outcome Sum $= 3: (1,2), (2,1) ightarrow 2$ outcomes Sum $= 5: (1,4), (2,3), (3,2), (4,1) ightarrow 4$ outcomes Sum $= 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) ightarrow 6$ outcomes Total outcomes for $A \cap \overline{B} = 1 + 2 + 4 + 6 = 13$. Therefore,$P(A \cap \overline{B}) = \frac{13}{36}$.

Two dice are thrown and the sum of the numbers appeared on the dice is noted. If $A$ is the event of getting a prime number as their sum and $B$ is the event of getting a number greater than $8$ as their sum,then $P(A \cap \overline{B})=$

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