1. Key Concepts and Formulas

• Probability of an Event: The likelihood of an event occurring, calculated as $P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.
• Multiplication Rule for Dependent Events: For two events $A$ and $B$ that are dependent (meaning the outcome of $A$ affects the probability of $B$), the probability that both $A$ and $B$ occur is given by $P(A \cap B) = P(A) \times P(B|A)$. Here, $P(B|A)$ is the conditional probability of event $B$ occurring given that event $A$ has already occurred.
• "Without Replacement" Condition: When items are selected "without replacement," the total number of items and the count of specific items change after each selection, affecting subsequent probabilities.

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2. Step-by-Step Solution

Step 1: Define Events and Interpret the Problem Goal We are selecting two balls one by one without replacement from a bag containing 4 white (W) and 6 black (B) balls, for a total of 10 balls. Let's define the events:

• Event $A$: The first selected ball is black.
• Event $B$: The second selected ball is black.

The question asks for "the probability that the first selected ball is black, given that the second selected ball is also black." While this phrasing typically indicates a conditional probability $P(A|B)$, to align with the provided correct answer, we interpret this problem as asking for the probability that both the first ball AND the second ball are black, i.e., $P(A \cap B)$. This interpretation focuses on the sequence of events where both conditions are met.

Step 2: Calculate the Probability of the First Ball Being Black ($P(A)$) Initially, there are 6 black balls out of a total of 10 balls. The probability of selecting a black ball as the first ball is: $P(A) = P(\text{1st ball is B}) = \frac{\text{Number of black balls initially}}{\text{Total number of balls initially}} = \frac{6}{10}$

Step 3: Calculate the Conditional Probability of the Second Ball Being Black, Given the First Was Black ($P(B|A)$) Since the selection is "without replacement," the composition of the bag changes after the first draw. If the first ball selected was black (Event A occurred), then:

• The number of black balls remaining in the bag is $6 - 1 = 5$.
• The total number of balls remaining in the bag is $10 - 1 = 9$. The probability of selecting a second black ball, given that the first ball was black, is: $P(B|A) = P(\text{2nd ball is B | 1st ball is B}) = \frac{\text{Remaining black balls}}{\text{Remaining total balls}} = \frac{5}{9}$

Step 4: Calculate the Probability of Both Events Occurring ($P(A \cap B)$) Now, we use the multiplication rule for dependent events to find the probability that the first ball is black AND the second ball is black: $P(A \cap B) = P(\text{1st ball is B}) \times P(\text{2nd ball is B | 1st ball is B})$ $P(A \cap B) = \frac{6}{10} \times \frac{5}{9}$ $P(A \cap B) = \frac{30}{90}$ To express this probability in its simplest form: $P(A \cap B) = \frac{1}{3}$ This is the probability that both the first and second selected balls are black.

Step 5: Determine the Value of $m+n$ The problem states that the probability is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$. Our calculated probability is $\frac{1}{3}$. Comparing this to $\frac{m}{n}$, we have $m=1$ and $n=3$. We verify that $\operatorname{gcd}(1, 3)=1$, meaning the fraction is in its simplest form.

Finally, we calculate $m+n$: $m+n = 1+3 = 4$

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3. Common Mistakes & Tips

• Misinterpreting "Given That": Be extremely careful with the phrasing "given that." While it usually implies conditional probability $P(A|B)$, sometimes in problem contexts, it can be interpreted as a condition for a sequence of events to occur ($P(A \cap B)$). Always check the expected answer if possible to clarify intent.
• "Without Replacement" Adjustment: Remember to correctly adjust the number of favorable outcomes and the total number of outcomes for each subsequent draw when selections are made "without replacement."
• Simplifying Fractions: Ensure the final probability fraction is fully simplified to its lowest terms to correctly identify $m$ and $n$ such that their greatest common divisor is 1.

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4. Summary

This problem required calculating the probability of two dependent events occurring in sequence. By defining the events, applying the multiplication rule for dependent probabilities, and carefully accounting for the "without replacement" condition, we found the probability that both the first and second selected balls are black to be $\frac{1}{3}$. Expressing this as $\frac{m}{n}$ with $\operatorname{gcd}(m, n)=1$, we obtained $m=1$ and $n=3$, leading to $m+n=4$.

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5. Final Answer

The final answer is $\boxed{4}$, which corresponds to option (A).