Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball i

Q: Wait, why do we need to consider two cases? Can't we just calculate the probability of picking a white ball from the combined pool of both bags?

That is a common trap! The problem is sequential. The composition of Bag A changes *depending* on what was transferred from Bag B. You cannot treat the bags as a single combined pool because the transfer is a specific, dependent event. You must analyze the two mutually exclusive scenarios separately.

Q: Does the order of events matter? What if we drew from Bag A first?

Yes, absolutely. The problem explicitly states the transfer happens *before* the draw. If you were to draw first, the composition of Bag A would be different, and the probability would change. Always follow the chronological order of events described in the problem.

Q: How do I know if the events are mutually exclusive?

Events are mutually exclusive if they cannot happen at the same time. In this case, the transferred ball is either white or black—it cannot be both. Therefore, the event 'transferring a white ball' and 'transferring a black ball' are mutually exclusive and exhaustive, which is exactly why the Law of Total Probability applies perfectly here.

Question

Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A. Then a ball is randomly drawn from the bag A. If the probability, that the ball drawn is white, is $\frac{p}{q}$, $	ext{gcd}(p, q) = 1$, then $p + q$ is equal to :

Options:

(A) 23

(B) 24

(C) 21

(D) 22

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a room with two bags, Bag A and Bag B. Bag A holds $9$ white balls and $8$ black balls, totaling $17$ balls. Bag B holds $6$ white balls and $4$ black balls, totaling $10$ balls.

You are tasked with a simple experiment: pick one ball from Bag B, drop it into Bag A, and then draw a ball from Bag A. We must determine the probability that the ball drawn from Bag A is white.

### The Two Universes

Because we do not know which ball was transferred, we must consider two mutually exclusive scenarios, or "universes." Let $E_1$ be the event that a white ball is transferred, and $E_2$ be the event that a black ball is transferred.

The probability of picking a white ball from Bag B is:
$$P(E_1) = \frac{6}{10} = \frac{3}{5}$$

Conversely, the probability of picking a black ball from Bag B is:
$$P(E_2) = \frac{4}{10} = \frac{2}{5}$$

### The Transformation of Bag A

Bag A originally contained $17$ balls. After the transfer, it contains $18$ balls.

If we are in Universe $1$ (a white ball was added), Bag A now contains $10$ white balls and $8$ black balls. The conditional probability of drawing a white ball is:
$$P(W|E_1) = \frac{10}{18} = \frac{5}{9}$$

If we are in Universe $2$ (a black ball was added), Bag A now contains $9$ white balls and $9$ black balls. The conditional probability of drawing a white ball is:
$$P(W|E_2) = \frac{9}{18} = \frac{1}{2}$$

### The Synthesis: Law of Total Probability

To find the final probability $P(W)$, we use the Law of Total Probability, which calculates the weighted sum of these outcomes:
$$P(W) = P(E_1) \cdot P(W|E_1) + P(E_2) \cdot P(W|E_2)$$

Substituting our values into the equation:
$$P(W) = \left( \frac{3}{5} \cdot \frac{5}{9} ight) + \left( \frac{2}{5} \cdot \frac{1}{2} ight)$$

Simplifying the terms, we find:
$$P(W) = \frac{1}{3} + \frac{1}{5}$$

Finding a common denominator of $15$, we obtain:
$$P(W) = \frac{5 + 3}{15} = \frac{8}{15}$$

### Final Calculation

We are given that the probability is $\frac{p}{q}$ where $	ext{gcd}(p, q) = 1$. Our result is $\frac{8}{15}$.

Since $8$ and $15$ are co-prime, we identify $p = 8$ and $q = 15$. The final sum is:
$$p + q = 8 + 15 = \mathbf{23}$$

Select Answer:

Visualized Solution

Initial State of Bags

The Transfer and Draw Process

Mutually Exclusive Transfer Events

Probability of Transferring a White Ball ( $E_{1}$ )

Drawing White after $E_{1}$

Probability of Transferring a Black Ball ( $E_{2}$ )

Drawing White after $E_{2}$

Law of Total Probability

Substituting the Probabilities

Simplifying the Terms

Calculating Final Probability

Comparing with $\frac{p}{q}$

Finding $p + q$

The Sigma Insight: Total Probability Theorem

Analyzing the Setup

The Two Universes

The Transformation of Bag A

The Synthesis

Final Calculation

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