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.

Explanation

Let E: 4 appears on the third toss

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

`P(E)=frac{1}{6}`

`P(F)=frac{1}{6} times frac{1}{6}=frac{1}{36}`

Now,

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

Therefore, required probability, `P(E | F)=frac{P(E cap F)}{P(F)}=frac{1}{6}`