Question

Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability that the ball drawn is white is \(\frac{29}{45}\), then n is equal to:

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

In probability problems involving multiple stages, apply conditional probability rules carefully to avoid errors.

Answer 1

The Correct Option is D

Step 1: Probability Calculation.
Probability of choosing a white ball from Bag 1 and adding it to Bag 2: \[ P(W \text{ from Bag 1}) = \frac{4}{9} \] Probability of choosing a black ball from Bag 1 and adding it to Bag 2: \[ P(B \text{ from Bag 1}) = \frac{5}{9} \] Now, probability of choosing a white ball from Bag 2: \[ P(W \text{ from Bag 2}) = \frac{n + 1}{n + 4} \times \frac{4}{9} + \frac{n}{n + 4} \times \frac{5}{9} = \frac{29}{45} \] Cross multiplying and simplifying, we find: \[ n = 6 \]

Answer 2

Step 1: Understand the problem.
Bag 1 contains 4 white and 5 black balls.
Bag 2 contains n white and 3 black balls.
A ball is transferred from Bag 1 to Bag 2, and then a ball is drawn from Bag 2.
The probability that the drawn ball is white is given as \( \frac{29}{45} \).

Step 2: Define events.
Let:
E₁ = Event that a white ball is transferred from Bag 1.
E₂ = Event that a black ball is transferred from Bag 1.

Probability of transferring a white ball from Bag 1:
\[ P(E₁) = \frac{4}{9}, \quad P(E₂) = \frac{5}{9}. \]

Step 3: Calculate probability of drawing a white ball from Bag 2.
Case 1: If a white ball is transferred, Bag 2 will have \((n + 1)\) white and 3 black balls.
\[ P(\text{White from Bag 2} | E₁) = \frac{n + 1}{(n + 1) + 3} = \frac{n + 1}{n + 4}. \]

Case 2: If a black ball is transferred, Bag 2 will have \(n\) white and \((3 + 1) = 4\) black balls.
\[ P(\text{White from Bag 2} | E₂) = \frac{n}{n + 4}. \]

Step 4: Total probability.
\[ P(\text{White}) = P(E₁) \times P(\text{White | E₁}) + P(E₂) \times P(\text{White | E₂}) \] \[ \frac{29}{45} = \frac{4}{9} \times \frac{n + 1}{n + 4} + \frac{5}{9} \times \frac{n}{n + 4}. \] Simplify numerator:
\[ \frac{29}{45} = \frac{1}{9(n + 4)} [4(n + 1) + 5n] = \frac{1}{9(n + 4)} (9n + 4). \] Simplify further:
\[ \frac{29}{45} = \frac{9n + 4}{9(n + 4)}. \] Multiply both sides by \(9(n + 4)\):
\[ \frac{29}{45} \times 9(n + 4) = 9n + 4. \] \[ \frac{29}{5}(n + 4) = 9n + 4. \] Multiply by 5:
\[ 29(n + 4) = 45n + 20. \] \[ 29n + 116 = 45n + 20. \] \[ 96 = 16n \Rightarrow n = 6. \]

Final Answer:
\[ \boxed{n = 6} \]