Question

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 $A$

• A. $9/25$
• B. $3/5$
• C. $27/125$
• D. $1/3$

Answer 1

The correct option is A $9/25$

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) = $\frac{1/8}{25/72}=9/25$