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’.

Explanation

When dice is thrown, number of observations in the sample space `=6×6=36`

Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers
appearing on throwing the two dice are different.

`∴A={(1,3),(2,2),(3,1)}`

`B={(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)}`

`A∩B={(1,3),(3,1)}`

`∴P(B)=(30)/(36)​=5/6`​ and
`P(A∩B)=2/(36)​=1/(18)​`

Let `P(A//B)` represent the probability that the sum of the numbers on the dice is `4`,
given that the two numbers appearing on throwing the two dice are different.

`P(A//B)=P(B)P(AnnB)​=(1/(18))/(5/6)​​`

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