Question

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is :

• A. \(\frac{4}{9}\)
• B. \(\frac{5}{18}\)
• C. \(\frac{1}{6}\)
• D. \(\frac{3}{10}\)

Answer 1

The Correct Option is B

The given problem involves probability with conditional events and can be solved using the concept of conditional probability.

First, let's define the scenario:

• Bag I contains 3 red, 4 black, and 3 white balls, totaling 10 balls.
• Bag II contains 2 red, 5 black, and 2 white balls, totaling 9 balls initially.

Let's consider the events:

• \(A\): The event that a red ball is transferred from Bag I to Bag II.
• \(B\): The event that the ball drawn from Bag II is black.

We need to find \(P(A \mid B)\), the probability that a red ball was transferred given that a black ball is drawn.

By Bayes' Theorem, \(P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}\).

Step 1: Calculate \(P(A)\)

The probability of transferring a red ball from Bag I, \(P(A)\), is:

P(A) = \frac{\text{Number of red balls in Bag I}}{\text{Total number of balls in Bag I}} = \frac{3}{10}

Step 2: Calculate \(P(B \mid A)\)

Given that a red ball is transferred to Bag II, the composition of Bag II becomes:

• 3 red balls,
• 5 black balls,
• 2 white balls.

The probability of drawing a black ball from Bag II in this case is:

P(B \mid A) = \frac{5}{10} = \frac{1}{2}

Step 3: Calculate \(P(B)\)

\(P(B)\) is the probability of drawing a black ball considering both possibilities (transferred ball is red, black, or white):

P(B) = P(B \mid A) \cdot P(A) + P(B \mid A') \cdot P(A')

• Where \(A'\): Event that a non-red ball is transferred (either black or white).

For \(A'\):

P(A') = 1 - P(A) = \frac{7}{10}

If a black ball is transferred:

• 5 red balls,
• 6 black balls,
• 2 white balls in Bag II

P(B \mid A', \text{transfer black}) = \frac{6}{11}

If a white ball is transferred:

• 2 red balls,
• 5 black balls,
• 3 white balls in Bag II

P(B \mid A', \text{transfer white}) = \frac{5}{11}

P(A'\text{ and transfer black}) = \frac{4}{10} and P(A'\text{ and transfer white}) = \frac{3}{10}

So,

P(B) = \frac{1}{2} \cdot \frac{3}{10} + \frac{6}{11} \cdot \frac{2}{5} + \frac{5}{11} \cdot \frac{3}{10}

= \frac{3}{20} + \frac{12}{55} + \frac{15}{110} = \frac{51}{110}

Step 4: Calculate \(P(A \mid B)\)

P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)} = \frac{\frac{1}{2} \cdot \frac{3}{10}}{\frac{51}{110}}

= \frac{3}{20} \times \frac{110}{51} = \frac{5}{18}

Therefore, the probability that the transferred ball was red, given that a black ball was drawn from Bag II, is \(\frac{5}{18}\).

Answer 2

Let E→Ball drawn from Bag II is black.
E_R→Bag I to Bag II red ball transferred.
E_B→Bag I to Bag II black ball transferred.
E_W→Bag I to Bag II white ball transferred.
 

\(P(\frac{E_R}{E})=\frac{P(\frac{E}{ER})⋅P(ER) }{ P(\frac{E}{ER})P(ER)+P(\frac{E}{EB})⋅P(EB)+P(\frac{E}{EW})P(EW)}\)

Here,
P(ER)=\(\frac{3}{10}\), P(EB)=\(\frac{4}{10}\), P(E_W)=\(\frac{3}{10}\)
and
 \(P(\frac{E}{E_R})=\frac{5}{10},P(\frac{E}{E_B})=\frac{5}{10},P(\frac{E}{E_W})=\frac{5}{10}\)

\(∴P(\frac{E_R}{E})=\frac{\frac{15}{100} }{\frac{ 15}{100}+\frac{24}{100}+\frac{15}{100}}\)
=\(\frac{15}{54}\)
\(=\frac{5}{18}\)
So, the correct option is (B): \(\frac{5}{18}\)