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(A) The total number of outcomes when three dice are thrown is $6^3 = 216$. Let $S$ be the sum of the numbers on the three dice. The possible values f

Vedclass · Accepted Answer

$\frac{35}{108}$

Vedclass · Answer

(A) The total number of outcomes when three dice are thrown is $6^3 = 216$. Let $S$ be the sum of the numbers on the three dice. The possible values for $S$ range from $3$ to $18$. Event $A$ is the sum $S > 14$,i.e.,$S \in \{15, 16, 17, 18\}$. The number of ways to get these sums are: $S=15: (6,6,3) 	imes 3, (6,5,4) 	imes 6, (5,5,5) 	imes 1 = 10$ ways. $S=16: (6,6,4) 	imes 3, (6,5,5) 	imes 3 = 6$ ways. $S=17: (6,6,5) 	imes 3 = 3$ ways. $S=18: (6,6,6) 	imes 1 = 1$ way. Total outcomes for $A = 10 + 6 + 3 + 1 = 20$. Event $B$ is the sum $S$ being a multiple of $3$,i.e.,$S \in \{3, 6, 9, 12, 15, 18\}$. Outcomes for $S=15$ are $10$ and for $S=18$ are $1$. Thus,$A \cap B$ contains outcomes where $S=15$ or $S=18$,so $n(A \cap B) = 10 + 1 = 11$. $P(A \cap \overline{B}) = P(A) - P(A \cap B) = \frac{20}{216} - \frac{11}{216} = \frac{9}{216}$. Now,$P(\overline{A} \cap B) = P(B) - P(A \cap B)$. Number of outcomes for $B$: $S=3: 1$,$S=6: 10$,$S=9: 25$,$S=12: 25$,$S=15: 10$,$S=18: 1$. Total $n(B) = 1 + 10 + 25 + 25 + 10 + 1 = 72$. $P(\overline{A} \cap B) = \frac{72}{216} - \frac{11}{216} = \frac{61}{216}$. Finally,$P(A \cap \overline{B}) + P(\overline{A} \cap B) = \frac{9}{216} + \frac{61}{216} = \frac{70}{216} = \frac{35}{108}$.

Three dice are thrown simultaneously and the sum of the numbers appeared on them is noted. If $A$ is the event of getting a sum greater than $14$ and $B$ is the event of getting a sum which is a multiple of $3$,then $P(A \cap \overline{B}) + P(\overline{A} \cap B) = $

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