Question

Determine P(E|F): A dice is thrown three times.
E: 4 appears on the third toss,
F: 6 and 5 appears respectively on first two tosses.

Answer 1

If a die is thrown three times, then the number of elements in the sample space will be:
6 × 6 × 6 = 216

E = (1,2,3,4,5,6) x (1,2,3,4,5,6) x (4)
F = (6) x (5) x (1,2,3,4,5,6)
⇒n(F) = 1x1x6 = 6

$P(F)=\frac {n(F)}{n(S)}$ =$\frac {6}{216 }$

∴E∩F = (6,5,4) ⇒ nE∩F = 1

∴$\frac {P(E∩F)}{n(S)} =\frac {1}{216}$

And, $P(E|F)=\frac {P(E∩F)}{P(F)} =\frac {1/216}{6/216 }=\frac 16$