# Suppose a girl thrown a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3,4,5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw 3,4,5 or 6 with the die?

Given:

Let us assume D1, D2 and A be the events as follows:

D1 = Throwing die and getting 1 or 2

D2 = Throwing die and getting 3,4,5 or 6

A = getting exactly one tail

P(A|D1) = P(getting 1 tail on getting 1 or 2 from die)

P(A|D2) = P(Getting 1 tail on getting 3,4,5 or 6 from die)

Now we find

P(D2|A) = P(The tail we get here come after getting 3,4,5 or 6 from die)

Using Baye’s theorem:

The required probability is .

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