Let event B is "5 appears on the first dice".
`thereforeB={(5,1),(5,2),(5,3),(5,4).(5,5),(5,6)}`
Also, let event A is "sum of the number appearing is 11".
`thereforeA={(5,6),(6,5)}`
` thereforeA fnnB=={(5,6)}`
Therefore, `P(A//b)=(P(AnnB))/(n(B))=1/6`