A coin is tossed twice if the coin shows head it is tossed again but if it shows a tail then a die is tossed. If 8 possible outcomes are equally likely. Find the probability that the die shows a number greater than 4, if it is known that the first throw of the coin results in a tail

Here Sample space S = { HH, HT, T1, T2, T3, T4, T5, T6 }

Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then,

 A = { T5, T6 }

 B = { T1, T2, T3, T4, T5, T6 }

Therefore, Required probability =  $P\frac{A}{B}=\frac{PA\cap B}{PB}=\frac{\frac{2}{8}}{\frac{6}{8}}=\frac{1}{3}$