Solution:

In the sample space, there are 216 outcomes, with each element of the sample space having three entries and taking the form

$(x, y, z)$ where $\1 leq x, y, z leq 6$.

Considering the event, E: 4 appears on the third toss

$$\Rightarrow \mathrm{E}=\left\{\begin{array}{l}

(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) \end{array}\right\}$$

Now the event, F: 6 and 5 appears respectively on first two tosses

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

$Rightarrow E cap F={(6,5,4)}$

So, $P(E)=frac{36}{216}, P(F)=frac{6}{216}, P(E cap F)=frac{1}{216}$

Now, we know that by definition of conditional probability, $P(E mid F)=frac{P(E cap F)}{P(F)}$

Now by substituting the values we get

$Rightarrow mathrm{P}(mathrm{E} mid mathrm{F})=frac{1 / 216}{6 / 216}=frac{1}{6}$

$Rightarrow mathrm{P}(mathrm{E} mid mathrm{F})=frac{1}{6}$