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
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⇒ P(A|D1) = P(getting 1 tail on getting 1 or 2 from die)
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⇒ P(A|D2) = P(Getting 1 tail on getting 3,4,5 or 6 from die)
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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:
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∴ The required probability is .