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: Tree diagram
Branch on what moves. White (3/8) then white in B (5/8). Black (5/8) then white in B (4/8).

Step 2: Multiply along branches
$\frac{3 \cdot 5}{64} = \frac{15}{64}$ and $\frac{5 \cdot 4}{64} = \frac{20}{64}$.

Step 3: Add
The two ways to get a white ball add up to $\frac{35}{64}$.

Step 4: Sanity check
Bag B started with $\frac{4}{7} = 0.571$ white. Most balls in A are black, so adding one ball from A lowers the white fraction a little, to $\frac{35}{64} = 0.547$. This is consistent with the result.

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