Question

Bag \( P \) contains 3 white, 2 red, 5 blue balls and bag \( Q \) contains 2 white, 3 red, 5 blue balls. A ball is chosen at random from \( P \) and is placed in \( Q \). If a ball is chosen from bag \( Q \) at random, then the probability that it is a red ball is:

• A. \( \frac{9}{50} \)
• B. \( \frac{13}{45} \)
• C. \( \frac{16}{55} \)
• D. \( \frac{12}{35} \)

Hint:

To calculate probabilities involving multiple stages, use the law of total probability and consider the outcomes for each possible event.

Answer 1

The Correct Option is C

We are given two bags: - Bag \( P \) contains:
- 3 white balls
- 2 red balls
- 5 blue balls
- Bag \( Q \) contains:
- 2 white balls
- 3 red balls
- 5 blue balls
A ball is first chosen randomly from Bag \( P \) and placed into Bag \( Q \). Then, a ball is randomly selected from Bag \( Q \). We need to determine the probability that the selected ball from Bag \( Q \) is red. 

Step 1: Define Probabilities for Transfer from \( P \) Since a ball is chosen at random from Bag \( P \): \[ P(\text{white}) = \frac{3}{10}, \quad P(\text{red}) = \frac{2}{10}, \quad P(\text{blue}) = \frac{5}{10} \] 

 Step 2: Update Bag \( Q \) Based on Transfer 1. If a white ball is transferred: - Bag \( Q \) now has 3 white, 3 red, 5 blue. - Probability of selecting a red ball from \( Q \): \[ P(R | W) = \frac{3}{(2+1+3+5)} = \frac{3}{11} \] 2. If a red ball is transferred:
- Bag \( Q \) now has 2 white, 4 red, 5 blue.
- Probability of selecting a red ball from \( Q \): \[ P(R | R) = \frac{4}{(2+3+1+5)} = \frac{4}{11} \] 3. If a blue ball is transferred: - Bag \( Q \) now has 2 white, 3 red, 6 blue. - Probability of selecting a red ball from \( Q \): \[ P(R | B) = \frac{3}{(2+3+5+1)} = \frac{3}{11} \] 

Step 3: Compute Total Probability Using Law of Total Probability \[ P(R) = P(W) P(R | W) + P(R) P(R | R) + P(B) P(R | B) \] \[ = \left(\frac{3}{10} \times \frac{3}{11} \right) + \left(\frac{2}{10} \times \frac{4}{11} \right) + \left(\frac{5}{10} \times \frac{3}{11} \right) \] \[ = \frac{9}{110} + \frac{8}{110} + \frac{15}{110} \] \[ = \frac{32}{110} = \frac{16}{55} \] 

Final Answer: \(\boxed{\frac{16}{55}}\)