Suppose 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 once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Given: : let E_{1} be the event that the outcome on the die is 5 or 6, E_{2} be the event that the outcome on the die is 1, 2, 3 or 4 and A be the event getting exactly head.

Then

And

As in throwing a coin three times we get 8 possibilities.

{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

⇒ P(A|E_{1}) = P (obtaining exactly one head by tossing the coin three times if she get 5 or 6)

And P(A|E_{2}) = P (obtaining exactly one head by tossing the coin three times if she get 1,2 ,3 or 4)

Now the probability that the girl threw 1, 2, 3 or 4 with a die, being given that she obtained exactly one head, is P(E_{2}|A).

By using bayes’ theorem, we have:

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