A man has 3 coins A, B & C. A is fair coin. B is biased such that the probability of occurring head on it is 2/3. C is also biased with the probability of occurring head as 1/3. If one coin is selected and tossed three times, giving two heads and one tail, find the probability that the chosen coin was A

Engineering

Mathematics

Conditional Probability

Question

9/25

3/5

27/125

1/3

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Outcome HHT
Coin A
$P (HHT)_{A} = (\frac{1}{2})^{3} = 1 / 8$
Coin B
$P (HHT)_{B} = (\frac{2}{3})^{2} . \frac{1}{3} = 4 / 27$
Coin C
$P (HHT)_{C} = (\frac{1}{3})^{2} . \frac{2}{3} = 2 / 27$ .
P(HHT)_total = 25/72
P(chosen coin was A) = $\frac{1 / 8}{25 / 72} = 9 / 25$

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