Question

Bag A contains 3 red and 5 black balls, bag B contains 5 red and 3 black balls and bag C contains 4 red and 4 black balls. A bag is chosen randomly and a ball is drawn randomly from the chosen bag. If the ball drawn is found to be black, then the probability that it is drawn from bag B is

• A. \(\frac{7}{12}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{5}{12}\)
• D. \(\frac{1}{3}\)

Hint:

Whenever a question asks “given that some event has already occurred” and you must identify the original source or cause, immediately think of Bayes’ theorem. It is one of the most frequently tested probability concepts.

Answer 1

The Correct Option is B

Concept: This problem is a direct application of Bayes’ Theorem, which is used when we need to determine the probability of a cause after observing an event. Bayes’ theorem is written as: \[ P(A_i|E)=\frac{P(A_i)\cdot P(E|A_i)}{\sum P(A_j)\cdot P(E|A_j)} \] where:
• \(A_i\) represents possible causes.
• \(E\) represents the observed event.
• \(P(A_i|E)\) is called posterior probability.
• \(P(E|A_i)\) is conditional probability. In this question:
• Three bags represent possible sources.
• A black ball drawn is the observed event.
• We need to determine probability that black ball came from bag B. Thus Bayes theorem is the most suitable approach.

Step 1: Define the events involved in the problem.
Let \[ A_1=\text{Selection of Bag A} \] \[ A_2=\text{Selection of Bag B} \] \[ A_3=\text{Selection of Bag C} \] Let event \[ E=\text{Drawing a black ball} \] Since one bag is chosen randomly among three bags, each bag has equal probability. Therefore \[ P(A_1)=P(A_2)=P(A_3)=\frac13 \]

Step 2: Find probability of drawing a black ball from each bag.
Bag A contains: \[ 3\text{ red},5\text{ black} \] Total balls: \[ 8 \] Thus probability of black from bag A is \[ P(E|A_1)=\frac58 \] Bag B contains: \[ 5\text{ red},3\text{ black} \] Thus probability of black from bag B is \[ P(E|A_2)=\frac38 \] Bag C contains: \[ 4\text{ red},4\text{ black} \] Thus probability of black from bag C is \[ P(E|A_3)=\frac48=\frac12 \]

Step 3: Apply Bayes theorem formula.
We need probability that selected bag was B given black ball is observed. Thus \[ P(A_2|E)= \frac{P(A_2)\times P(E|A_2)} {P(A_1)P(E|A_1)+P(A_2)P(E|A_2)+P(A_3)P(E|A_3)} \] Substituting values \[ P(A_2|E)= \frac{\left(\frac13\right)\left(\frac38\right)} {\left(\frac13\right)\left(\frac58\right)+\left(\frac13\right)\left(\frac38\right)+\left(\frac13\right)\left(\frac12\right)} \]

Step 4: Simplify numerator and denominator carefully.
Numerator becomes \[ \frac{1}{8} \] Denominator becomes \[ \frac{5}{24}+\frac{3}{24}+\frac{4}{24} \] \[ =\frac{12}{24} \] \[ =\frac12 \] Thus \[ P(A_2|E)=\frac{\frac18}{\frac12} \] \[ =\frac18\times2 \] \[ =\frac14 \]

Step 5: Write final answer.
Hence probability that black ball was drawn from bag B is \[ \boxed{\frac14} \] This corresponds to option (2).