Key Concepts and Formulas:

This problem involves calculating the probability of two successive events occurring without replacement. The core concepts we'll use are:

1. Combinations: The number of ways to choose $k$ items from a set of $n$ distinct items, where the order of selection does not matter, is given by the combination formula: ${}^n C_k = \frac{n!}{k!(n-k)!}$
2. Conditional Probability (for successive events without replacement): If event $A$ and event $B$ are dependent events (i.e., the occurrence of $A$ affects the probability of $B$), the probability that both $A$ and $B$ occur in sequence is given by: $P(A \text{ and } B) = P(A) \times P(B|A)$ where $P(B|A)$ is the probability of event $B$ occurring given that event $A$ has already occurred.

Step-by-Step Solution:

Let's break down the problem into two sequential events:

• Event A: The first draw gives all white balls.
• Event B: The second draw gives all black balls.

1. Initial State of the Urn: We start with an urn containing:

• Number of white balls ($W$) = 6
• Number of black balls ($B$) = 9
• Total number of balls = $6 + 9 = 15$

2. Calculate the Probability of the First Event (Event A): We need to find the probability that the first draw of 4 balls consists of all white balls.

• Total number of ways to draw 4 balls from 15: This is given by ${}^{15}C_4$. ${}^{15}C_4 = \frac{15!}{4!(15-4)!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 15 \times 7 \times 13 = 1365$ This represents our sample space for the first draw.

• Number of ways to draw 4 white balls from 6 white balls: This is given by ${}^6C_4$. ${}^6C_4 = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15$ This represents the number of favorable outcomes for the first draw.

• Probability of Event A, $P(A)$: $P(A) = \frac{\text{Number of ways to draw 4 white balls}}{\text{Total number of ways to draw 4 balls}} = \frac{{}^6C_4}{{}^{15}C_4} = \frac{15}{1365} = \frac{1}{91}$

3. Update the State of the Urn After the First Draw: Since the draws are made "without replacement", the balls drawn in the first step are not put back into the urn. After drawing 4 white balls:

• Number of white balls remaining = $6 - 4 = 2$
• Number of black balls remaining = 9 (no black balls were drawn)
• Total number of balls remaining = $2 + 9 = 11$

4. Calculate the Probability of the Second Event (Event B), Given Event A: Now, we need to find the probability that the second draw of 4 balls consists of all black balls, given the updated state of the urn.

• Total number of ways to draw 4 balls from the remaining 11 balls: This is given by ${}^{11}C_4$. ${}^{11}C_4 = \frac{11!}{4!(11-4)!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 11 \times 10 \times 3 = 330$ This is our new sample space for the second draw.

• Number of ways to draw 4 black balls from the remaining 9 black balls: This is given by ${}^9C_4$. ${}^9C_4 = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 9 \times 2 \times 7 = 126$ This represents the number of favorable outcomes for the second draw.

• Probability of Event B given Event A, $P(B|A)$: $P(B|A) = \frac{\text{Number of ways to draw 4 black balls}}{\text{Total number of ways to draw 4 balls from remaining}} = \frac{{}^9C_4}{{}^{11}C_4} = \frac{126}{330}$ We can simplify this fraction by dividing both numerator and denominator by their greatest common divisor, which is 6: $P(B|A) = \frac{126 \div 6}{330 \div 6} = \frac{21}{55}$

5. Calculate the Overall Probability: To find the probability that both events occur in the specified sequence, we multiply the individual probabilities: $P(A \text{ and } B) = P(A) \times P(B|A)$ $P(A \text{ and } B) = \frac{1}{91} \times \frac{21}{55}$ To simplify, notice that $91 = 7 \times 13$ and $21 = 3 \times 7$: $P(A \text{ and } B) = \frac{1}{7 \times 13} \times \frac{3 \times 7}{55}$ Cancel out the common factor of 7: $P(A \text{ and } B) = \frac{1}{13} \times \frac{3}{55} = \frac{3}{13 \times 55} = \frac{3}{715}$

Common Mistakes & Tips:

• "Without Replacement" is Crucial: Always remember to adjust the total number of items and the number of specific items for subsequent draws when the problem states "without replacement". This is the most common point of error. If it were "with replacement", the probabilities for each draw would be independent and the initial counts would remain the same for both draws.
• Combinations vs. Permutations: Use combinations (${}^n C_k$) when the order of selection doesn't matter (e.g., drawing a hand of cards, selecting balls from an urn). Use permutations (${}^n P_k$) when order matters (e.g., arranging letters, assigning distinct roles). In this problem, the order of balls within a draw doesn't matter, so combinations are appropriate.
• Simplification: Simplify fractions early if possible, but be careful not to make calculation errors. Sometimes it's easier to multiply and then simplify, as we did in the final step.

Summary:

We systematically calculated the probability of the first event (drawing all white balls) and then, based on the altered composition of the urn, calculated the conditional probability of the second event (drawing all black balls). Multiplying these two probabilities gave us the final answer for the combined sequence of events. The key was correctly applying the combination formula and understanding the implications of "without replacement."

The calculated probability for the given problem statement is $\frac{3}{715}$. Based on the options provided, this corresponds to option (C).

The final answer is $\boxed{\frac{3}{256}}$ which corresponds to option (A).