Question

Bag $A$ contains $3$ white and $4$ black balls. Bag $B$ contains $4$ white and $3$ black balls. Bag $C$ contains $2$ white and $5$ black balls. A bag is randomly selected and then a ball is randomly drawn from that bag. If the ball drawn was found to be white, then the probability that the ball is drawn from bag $C$ is

• A. $\dfrac{1}{6}$
• B. $\dfrac{2}{9}$
• C. $\dfrac{1}{4}$
• D. $\dfrac{2}{13}$

Hint:

Bayes' theorem: Posterior probability = $\dfrac{\text{Prior}\times\text{Likelihood}}{\text{Total Probability}}$.

Answer 1

The Correct Option is B

Step 1: Set the bag probabilities.
A bag is chosen at random from $A,B,C$, so each has probability $P(A)=P(B)=P(C)=\dfrac13.$
Step 2: White-ball chances from each bag.
Each bag holds $7$ balls. $P(W|A)=\dfrac37,$ $P(W|B)=\dfrac47,$ $P(W|C)=\dfrac27.$
Step 3: Total chance of white.
$P(W)=\dfrac13\left(\dfrac37+\dfrac47+\dfrac27\right)=\dfrac13\cdot\dfrac97=\dfrac{9}{21}=\dfrac37.$
Step 4: Apply Bayes' theorem.
We want the chance it came from $C$ given white: $P(C|W)=\dfrac{P(C)P(W|C)}{P(W)}.$
Step 5: Put in the numbers.
$P(C|W)=\dfrac{\frac13\cdot\frac27}{\frac37}=\dfrac{\frac2{21}}{\frac37}.$
Step 6: Simplify.
$\dfrac{2}{21}\div\dfrac37=\dfrac{2}{21}\cdot\dfrac{7}{3}=\dfrac{14}{63}=\dfrac29.$ \[ \boxed{\dfrac29} \]