A girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin two times and notes the number of heads obtained. If she obtained exactly two heads, what is the probability that she threw 1, 2, 3 or 4 with the die?

Let A₁ be when the girl gets 5 or 6 and hence tosses a coin 3 times.

Let A₂ be when the girl gets 1, 2, 3 or 4 and hence tosses a coin 2 times and A be when the girl gets exactly two heads.

⇒ P (A₁) = P (5 or 6) = P (5) + P (6)

⇒ P (A₂) = P (1, 2, 3 or 4) = P (1) + P (2) + P (3) + P (4)

And

Here n = 3;

sample space = {HHH, HTH, HHT, THH, HTT, THT, TTH, TTT}

And

Here n = 2;

Sample space = {HH, HT, TH, TT}

By Bayes’ Theorem,

Probability that the girl threw 1, 2, 3 or 4 with the die is 4/7.