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 $A$

Outcome HHT

Coin A

$P(HHT)_A = (\cfrac{1}{2})^3 = 1/8$

Coin B

$P(HHT)_B = (\cfrac{2}{3})^2.\cfrac{1}{3} = 4/27$

Coin C

$P(HHT)_C = (\cfrac{1}{3})^2.\cfrac{2}{3} = 2/27$.

$P(HHT)_{total} = 25/72$

P(chosen coin was A) = $\cfrac{1/8}{25/72} = 9/25$