Here,
`A = `Event that `6` and `5` appears on first two tosses.
It means third toss can have any of the six values.
`:. n(A) = 6`
Let `B = `Event that `4` appears on the third toss.
Now, `(AnnB) = {(6,5,4)}`
`=>n(AnnB) = 1`
`:.` Required probability , `P(B/A) = (n(AnnB) )/(n(A)) = 1/6.`