A die is thrown three times,

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

$\mathrm{E}=\left\{\begin{array}{l}(1,1,4),(1,2,4), \ldots(1,6,4) \\ (2,1,4),(2,2,4), \ldots(2,6,4) \\ (3,1,4),(3,2,4), \ldots(3,6,4) \\ (4,1,4),(4,2,4), \ldots(4,6,4) \\ (5,1,4),(5,2,4), \ldots(5,6,4) \\ (6,1,4),(6,2,4), \ldots(6,6,4)\end{array}\right\}$

$\mathrm{F}=\{(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)\}$

$\therefore \mathrm{E} \cap \mathrm{F}=\{(6,5,4)\}$

$P(F)=\frac{6}{216}$ and $P(E \cap F)=\frac{1}{216}$

$P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{6}$