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(A) When two dice are thrown,the total number of outcomes in the sample space is $6 	imes 6 = 36$.Let $A$ be the event that the sum of the numbers on

Vedclass · Accepted Answer

$\frac{1}{15}$

Vedclass · Answer

(A) When two dice are thrown,the total number of outcomes in the sample space is $6 	imes 6 = 36$.Let $A$ be the event that the sum of the numbers on the dice is $4$. The outcomes for $A$ are $\{(1,3), (2,2), (3,1)\}$.Let $B$ be the event that the two numbers appearing on the dice are different. The total number of outcomes where the numbers are the same is $6$ (i.e.,$\{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}$).Therefore,the number of outcomes in $B$ is $36 - 6 = 30$.We need to find $P(A|B) = \frac{P(A \cap B)}{P(B)}$.The intersection $A \cap B$ represents the event where the sum is $4$ $AND$ the numbers are different. From set $A$,the outcome $(2,2)$ has the same numbers,so $A \cap B = \{(1,3), (3,1)\}$.Thus,$n(A \cap B) = 2$ and $n(B) = 30$.The conditional probability is $P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{2}{30} = \frac{1}{15}$.

Given that the two numbers appearing on throwing two dice are different,find the probability of the event 'the sum of numbers on the dice is $4$.'

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