The sample space has 216 outcomes, where each element of the sample space has 3 entries and is of the form (x, y, z) where 1≤x,y,z≤6 . Here, E: 4 appears on the third toss⇒E=⎩⎨⎧​(1,1,4),(1,2,4),(1,3,4),(1,4,4),(1,5,4),(1,6,4),(2,1,4),(2,2,4),(2,3,4),(2,4,4),(2,5,4),(2,6,4)(3,1,4),(3,2,4),(3,3,4),(3,4,4),(3,5,4),(3,6,4)(4,1,4),(4,2,4),(4,3,4),(4,4,4),(4,5,4),(4,6,4),(5,1,4),(5,2,4),(5,3,4),(5,4,4),(5,5,4),(5,6,4),(6,1,4),(6,2,4),(6,3,4),(6,4,4),(6,5,4),(6,6,4)​⎭⎬⎫​ Now, F: 6 and 5 appears respectively on first two tosses|​⇒F={(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)}⇒E∩F={(6,5,4)} So, P(E)=21636​,P(F)=2166​,P(E∩F)=2161​​ Now, we know that by definition of conditional probability,P(E∣F)=P(F)P(E∩F)​ Now by substituting the values we get​⇒P(E∣F)=6/2161/216​=61​⇒P(E∣F)=61​​