Question

Bag A contains 3 white and 5 black balls while bag B contains 4 white and 3 black balls. A ball is selected at random from bag A and put in bag B. If a ball is now selected at random from bag B then the probability that this ball is white ball is...

• A. \(\frac{20}{64}\)
• B. \(\frac{31}{64}\)
• C. \(\frac{35}{64}\)
• D. \(\frac{32}{64}\)

Hint:

Use the total probability theorem over the colour of the ball transferred from bag A.

Answer 1

The Correct Option is C

Step 1: Understand the concept
The ball moved from A to B is either white or black. Bag B then has 8 balls, and we use total probability over the two cases.

Step 2: Case 1: a white ball moved
\(P = \frac{3}{8}\). Bag B now has 5 white and 3 black, so \(P(\text{white from B}) = \frac{5}{8}\). Contribution: \(\frac{3}{8}\cdot\frac{5}{8} = \frac{15}{64}\).

Step 3: Case 2: a black ball moved
\(P = \frac{5}{8}\). Bag B now has 4 white and 4 black, so \(P(\text{white from B}) = \frac{4}{8}\). Contribution: \(\frac{5}{8}\cdot\frac{4}{8} = \frac{20}{64}\).

Step 4: Add
\[ P = \frac{15}{64} + \frac{20}{64} = \frac{35}{64} \]
Option (C).

Final Answer:
The probability is 35/64. This is option (C). \[ \boxed{\text{(C) }\frac{35}{64}} \]