Key Concepts and Formulas

• Law of Total Probability: This fundamental principle allows us to calculate the overall probability of an event by considering all mutually exclusive and exhaustive conditions under which the event can occur. If $A_1, A_2, \dots, A_k$ are such conditions, and $E$ is an event, then $P(E) = \sum_{i=1}^k P(E|A_i)P(A_i)$.
• Conditional Probability: $P(E|A)$ is the probability of event $E$ occurring given that event $A$ has already occurred.
• Basic Probability Calculation: The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes.

Step-by-Step Solution

Step 1: Define Events and Initial Setup First, we clearly state the initial contents of each bag and define the key events involved in the experiment.

• Bag 1:

  • Number of white balls ($W_1$) = 4
  • Number of black balls ($B_1$) = 5
  • Total balls in Bag 1 = $4 + 5 = 9$

• Bag 2:

  • Number of white balls ($W_2$) = $n$
  • Number of black balls ($B_2$) = 3
  • Total balls in Bag 2 = $n + 3$

Now, let's define the events for the two stages of the experiment:

• Event $A_W$: A white ball is drawn from Bag 1 and transferred to Bag 2.
  • Why define this? This is one of the two possible outcomes of the first stage of the experiment, and it will change the composition of Bag 2.
• Event $A_B$: A black ball is drawn from Bag 1 and transferred to Bag 2.
  • Why define this? This is the other possible outcome of the first stage. Events $A_W$ and $A_B$ are mutually exclusive (you can't transfer both a white and a black ball at the same time) and exhaustive (you must transfer either a white or a black ball).
• Event $E$: The ball drawn randomly from Bag 2 (in the second stage) is white.
  • Why define this? This is the ultimate event whose probability we are interested in. We are given $P(E) = \frac{29}{45}$ and need to use this to find the value of $n$.

Step 2: Calculate Probabilities of the Transfer Events ($P(A_W)$ and $P(A_B)$) These are the probabilities of the initial conditions (transferring a white or black ball). They are calculated from the contents of Bag 1.

• Probability of transferring a white ball ($P(A_W)$): There are 4 white balls out of a total of 9 balls in Bag 1. $P(A_W) = \frac{\text{Number of white balls in Bag 1}}{\text{Total balls in Bag 1}} = \frac{4}{9}$

• Probability of transferring a black ball ($P(A_B)$): There are 5 black balls out of a total of 9 balls in Bag 1. $P(A_B) = \frac{\text{Number of black balls in Bag 1}}{\text{Total balls in Bag 1}} = \frac{5}{9}$

  • Self-check: Notice that $P(A_W) + P(A_B) = \frac{4}{9} + \frac{5}{9} = 1$. This confirms that these two events cover all possibilities for the first stage.

Step 3: Calculate Conditional Probabilities of Drawing a White Ball from Bag 2 ($P(E|A_W)$ and $P(E|A_B)$) These probabilities depend on how the composition of Bag 2 changes after a ball is transferred.

• Case 1: A white ball was transferred (Event $A_W$ occurred).

  • Why this case? If a white ball was transferred, the number of white balls in Bag 2 increases by one.
  • Composition of Bag 2 after transfer:
    • Number of white balls = $n + 1$ (original $n$ plus the transferred white ball)
    • Number of black balls = 3 (unchanged)
    • Total balls in Bag 2 = $(n+1) + 3 = n+4$
  • Conditional probability of drawing a white ball from this modified Bag 2 ($P(E|A_W)$): $P(E|A_W) = \frac{\text{Number of white balls in Bag 2}}{\text{Total balls in Bag 2}} = \frac{n+1}{n+4}$

• Case 2: A black ball was transferred (Event $A_B$ occurred).

  • Why this case? If a black ball was transferred, the number of black balls in Bag 2 increases by one.
  • Composition of Bag 2 after transfer:
    • Number of white balls = $n$ (unchanged)
    • Number of black balls = $3 + 1 = 4$ (original 3 plus the transferred black ball)
    • Total balls in Bag 2 = $n + 4$
  • Conditional probability of drawing a white ball from this modified Bag 2 ($P(E|A_B)$): $P(E|A_B) = \frac{\text{Number of white balls in Bag 2}}{\text{Total balls in Bag 2}} = \frac{n}{n+4}$
  • Common Mistake Alert: It's crucial to correctly update the total number of balls in Bag 2. In both cases, one ball is added, so the total becomes $(n+3)+1 = n+4$. Forgetting to update the total or incorrectly updating the number of white/black balls is a frequent error.

Step 4: Apply the Law of Total Probability and Solve for n Now, we combine the probabilities calculated in the previous steps using the Law of Total Probability formula: $P(E) = P(E|A_W)P(A_W) + P(E|A_B)P(A_B)$ We are given that $P(E) = \frac{29}{45}$. Let's substitute all the values: $\frac{29}{45} = \left(\frac{n+1}{n+4}\right) \left(\frac{4}{9}\right) + \left(\frac{n}{n+4}\right) \left(\frac{5}{9}\right)$ To simplify, combine the terms on the right-hand side. Notice they share a common denominator $9(n+4)$: $\frac{29}{45} = \frac{4(n+1)}{9(n+4)} + \frac{5n}{9(n+4)}$ $\frac{29}{45} = \frac{4(n+1) + 5n}{9(n+4)}$ Now, expand the numerator: $\frac{29}{45} = \frac{4n + 4 + 5n}{9(n+4)}$ $\frac{29}{45} = \frac{9n + 4}{9(n+4)}$ We can simplify by dividing both sides by $\frac{1}{9}$: $\frac{29}{5} = \frac{9n + 4}{n+4}$ Now, cross-multiply to solve for $n$: $29(n+4) = 5(9n+4)$ $29n + 116 = 45n + 20$ Gather $n$ terms on one side and constant terms on the other: $116 - 20 = 45n - 29n$ $96 = 16n$ Divide by 16 to find $n$: $n = \frac{96}{16}$ $n = 6$

Step 5: Verify the Answer Let's quickly check if $n=6$ yields the given probability. If $n=6$: $P(E) = \frac{9(6) + 4}{9(6+4)} = \frac{54+4}{9(10)} = \frac{58}{90} = \frac{29}{45}$. This matches the given probability, confirming our derived value of $n$.

Common Mistakes & Tips

• Forgetting to update total balls: When a ball is transferred, the total number of balls in the receiving bag changes. Always update the total number of balls in the bag for subsequent probability calculations.
• Incorrectly updating ball counts: Ensure you add the transferred ball to the correct color count in the receiving bag. If a white ball is transferred, only the white ball count increases.
• Algebraic errors: Be careful with algebraic manipulation when solving for $n$. Double-check expansion, combining like terms, and cross-multiplication.
• Understanding "Mutually Exclusive and Exhaustive": The initial conditions (transferring a white or black ball) must cover all possibilities and cannot happen simultaneously for the Law of Total Probability to be applied correctly.

Summary

This problem is a classic application of the Law of Total Probability. We systematically defined the initial conditions (transferring a white or black ball from Bag 1 to Bag 2) and their respective probabilities. Then, for each condition, we determined the conditional probability of drawing a white ball from Bag 2 after the transfer. By combining these using the Law of Total Probability, we set up an equation that allowed us to solve for $n$. The derived value of $n=6$ was verified against the given probability.

The final answer is $\boxed{\text{6}}$, which corresponds to option (B).