Question

A die is thrown three times, find the probability that $4$ appears on the third toss if it is given that $6$ and $5$ appear respectively on first two tosses.

Answer 1

Let E: 4 appears on the third toss

F: 6 and 5 appear respectively on the first two tosses.

$P\left(E\right)=\frac{1}{6}$

$P\left(F\right)=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$

Now,

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

Therefore, required probability , $P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{1}{6}$