Two balls are selected at random one by one without replacement from a bag containing 4 white and 6

Q: Wait, why can't I just say the probability is $6/10$ for the second ball? Isn't it just a random draw?

That is a classic trap! If we knew nothing about the first draw, the probability of the second ball being black would indeed be $6/10$ due to symmetry. However, the problem explicitly gives us information: we are *given* that the second ball is black. This condition changes the sample space and forces us to account for the dependency created by the 'without replacement' rule.

Q: Why do we need to calculate $P(W_1 \cap B_2)$? Doesn't the condition 'given the second is black' fix the outcome?

The condition 'given the second is black' tells us the *final* state, but it doesn't tell us *how* we got there. The second ball being black could have happened after a black ball was removed, or after a white ball was removed. We must include both possibilities in our total probability calculation to correctly weight the likelihood of the first event.

Q: Do I really need to use Bayes' Theorem here?

You don't need to memorize the formal Bayes' Theorem formula if you understand the logic of conditional probability. The definition $P(A|B) = \frac{P(A \cap B)}{P(B)}$ is sufficient. Think of it as restricting the universe of all possibilities to only those where the second ball is black, and then finding what fraction of that restricted universe started with a black ball.

Question

Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is $\frac{m}{n}$ where $\gcd(m,n)=1$, then $m+n$ is equal to:

Options:

(A) 4

(B) 14

(C) 13

(D) 11

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a bag containing $4$ white balls and $6$ black balls. You are about to embark on a two-step experiment: drawing two balls, one by one, without putting the first one back.

This simple act of 'without replacement' is the heartbeat of this problem. It introduces a memory into the system; the bag remembers what you took, and that memory dictates the future.

### The Anatomy of Conditional Probability

We are tasked with finding the probability that the first ball was black, given that the second ball is black. Let $B_1$ be the event that the first ball is black, and $B_2$ be the event that the second ball is black.

We seek $P(B_1 | B_2)$. By the fundamental definition of conditional probability, we know that:

$$P(B_1 | B_2) = \frac{P(B_1 \cap B_2)}{P(B_2)}$$

This formula is our map. The numerator, $P(B_1 \cap B_2)$, is the probability of the path where both balls are black. The denominator, $P(B_2)$, is the total probability of the second ball being black, regardless of what happened first.

### The Two Paths to the Second Black Ball

To find $P(B_2)$, we must account for every possible way to reach that outcome. There are two distinct paths: Path A, where we draw a black ball first, and Path B, where we draw a white ball first.

For Path A, the probability of drawing a black ball first is $P(B_1) = \frac{6}{10}$. Once that black ball is gone, the bag holds $9$ balls, $5$ of which are black. Thus, the probability of the second being black is $P(B_2 | B_1) = \frac{5}{9}$.

Multiplying these, we get:

$$P(B_1 \cap B_2) = \frac{6}{10} 	imes \frac{5}{9} = \frac{30}{90}$$

Now, consider Path B. The probability of drawing a white ball first is $P(W_1) = \frac{4}{10}$. If we remove a white ball, the bag still contains all $6$ black balls, but only $9$ balls in total.

So, $P(B_2 | W_1) = \frac{6}{9}$. Multiplying these gives:

$$P(W_1 \cap B_2) = \frac{4}{10} 	imes \frac{6}{9} = \frac{24}{90}$$

### The Grand Synthesis

The total probability of the second ball being black is the sum of these two mutually exclusive paths:

$$P(B_2) = P(B_1 \cap B_2) + P(W_1 \cap B_2) = \frac{30}{90} + \frac{24}{90} = \frac{54}{90}$$

Now, we return to our conditional probability formula:

$$P(B_1 | B_2) = \frac{30/90}{54/90}$$

The $90$ in the denominator cancels out beautifully, leaving us with $\frac{30}{54}$. Simplifying this fraction by dividing both numerator and denominator by $6$, we arrive at $\frac{5}{9}$.

The problem states this is $\frac{m}{n}$ with $\gcd(m, n) = 1$. Thus, $m = 5$ and $n = 9$. The final step is:

**$m + n = 5 + 9 = 14$**

Select Answer:

Visualized Solution

Visualizing the Bag

Defining the Goal

First Draw: Black Ball

Second Draw: Black after Black

Intersection: Both Black

First Draw: White Ball

Intersection: White then Black

Total Probability of Second Black

Substituting into Bayes' Formula

Simplifying the Fraction

Final Calculation

The Sigma Insight: Total Probability Theorem

Analyzing the Setup

The Anatomy of Conditional Probability

The Two Paths to the Second Black Ball

The Grand Synthesis

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