Consider the experiment of throwing a die. if a multiple of 3 comes tip. throw the die again and if any other number comes, toss a coin Find the conditional probability of the event the coin shows a tail, given that at least one die shows a 3.

Explanation

`{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(1, H),(1, T),(2, H),(2, T),(4, T),(4, H),(5, T),(5, H),(6, T),(6, H)}`

E: coin shows a tail

`P(E)=frac{5}{16}`

F : At least one die shows a 3

`P(F)=frac{6}{16}`

`P(E_{cap} F)=frac{0}{12}=0 `

`therefore P(E | F)=frac{P(E cap F)}{P(F)}=frac{0}{frac{6}{16}}=0`

`therefore` Event is called impossible event