Key Concepts and Formulas

1. Distribution of Distinct Items into Distinct Bins: When $n$ distinct items are distributed into $m$ distinct bins, each item has $m$ independent choices for placement. The total number of ways is $m^n$. This is the fundamental principle for calculating the total sample space in such problems.
2. Combinations ($^nC_r$): The number of ways to choose $r$ distinct items from a set of $n$ distinct items, where the order of selection does not matter, is given by the combination formula: $^nC_r = \frac{n!}{r!(n-r)!}$ This is used when selecting a subset of items for a specific condition.
3. Probability of an Event: The probability of an event $E$ is the ratio of the number of favorable outcomes to the total number of possible outcomes: $P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$

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Step-by-Step Solution

We are given 9 distinct balls and 4 distinct boxes ($B_1, B_2, B_3, B_4$). We need to find the probability that box $B_3$ contains exactly 3 balls. This probability is given in the form $k \left( \frac{3}{4} \right)^9$.

Step 1: Calculate the Total Number of Possible Outcomes ($N_{total}$)

• What we are doing: Determining all possible ways to distribute the 9 distinct balls among the 4 distinct boxes without any restrictions.
• Why this approach: Each of the 9 distinct balls can be placed into any of the 4 distinct boxes, independently of the other balls. Since the balls are distinct, swapping two balls between boxes results in a different outcome. Since boxes are distinct, placing a ball in $B_1$ is different from placing it in $B_2$. This is a direct application of the $m^n$ rule.
• Calculation:
  • Ball 1 has 4 choices ($B_1, B_2, B_3, B_4$).
  • Ball 2 has 4 choices.
  • ...
  • Ball 9 has 4 choices. Therefore, the total number of ways to distribute the 9 distinct balls is: $N_{total} = 4^9$

Step 2: Calculate the Number of Favorable Outcomes ($N_{favorable}$)

• What we are doing: Determining the number of ways such that box $B_3$ contains exactly 3 balls. This requires a two-stage process: first, selecting which 3 balls go into $B_3$, and second, distributing the remaining balls into the remaining boxes.

• Sub-step 2a: Choosing 3 balls for Box $B_3$

  • What we are doing: Selecting 3 specific balls out of the 9 distinct balls to be placed in box $B_3$.
  • Why this approach: Since the balls are distinct and the order of selection for $B_3$ does not matter (i.e., picking ball A then B then C is the same as picking B then C then A for $B_3$), we use combinations.
  • Calculation: The number of ways to choose 3 balls out of 9 distinct balls is: $^9C_3 = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$

• Sub-step 2b: Distributing the remaining balls

  • What we are doing: After 3 balls have been chosen for $B_3$, there are $9 - 3 = 6$ balls remaining. These 6 balls must not be placed in $B_3$, because $B_3$ must contain exactly 3 balls. Therefore, these 6 remaining distinct balls must be distributed among the other $4 - 1 = 3$ boxes ($B_1, B_2, B_4$).
  • Why this approach: Similar to Step 1, each of these 6 distinct balls can be placed into any of the 3 available distinct boxes independently. This is again an application of the $m^n$ rule, with $n=6$ balls and $m=3$ boxes.
  • Calculation: The number of ways to distribute the remaining 6 balls into the other 3 boxes is: $3^6 = 729$

• Total Favorable Outcomes: To get the total number of favorable outcomes, we multiply the results of Sub-step 2a and Sub-step 2b, as these are sequential and independent events in fulfilling the condition. $N_{favorable} = (^9C_3) \times (3^6) = 84 \times 729$

Step 3: Calculate the Probability ($P(E)$)

• What we are doing: Using the fundamental probability formula to find the probability of the event.
• Why this approach: This is the standard definition of probability for equally likely outcomes.
• Calculation: $P(E) = \frac{N_{favorable}}{N_{total}} = \frac{^9C_3 \times 3^6}{4^9}$ $P(E) = \frac{84 \times 3^6}{4^9}$

Step 4: Determine the Value of $k$

• What we are doing: Equating our calculated probability with the given form $k \left( \frac{3}{4} \right)^9$ to solve for $k$.
• Why this approach: The problem explicitly provides the probability in a specific format, and we need to find the unknown coefficient $k$.
• Calculation: $k \left( \frac{3}{4} \right)^9 = \frac{84 \times 3^6}{4^9}$ $k \frac{3^9}{4^9} = \frac{84 \times 3^6}{4^9}$ Multiplying both sides by $4^9$: $k \times 3^9 = 84 \times 3^6$ Dividing both sides by $3^9$: $k = \frac{84 \times 3^6}{3^9}$ Using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$: $k = 84 \times 3^{(6-9)} = 84 \times 3^{-3}$ $k = 84 \times \frac{1}{3^3} = 84 \times \frac{1}{27}$ Simplifying the fraction by dividing both numerator and denominator by 3: $k = \frac{84 \div 3}{27 \div 3} = \frac{28}{9}$

Step 5: Identify the set to which $k$ belongs

• What we are doing: Comparing the calculated value of $k$ with the given options, which are expressed as intervals.

• Why this approach: This is the final step to answer the multiple-choice question.

• Calculation: We have $k = \frac{28}{9}$. Let's approximate its value: $k \approx 3.111...$

  • (A) {x $\in$ R : |x $-$ 3| < 1}: $|x - 3| < 1 \implies -1 < x - 3 < 1 \implies 2 < x < 4$. Since $2 < 3.111... < 4$, $k$ lies in this set.

  • (B) {x $\in$ R : |x $-$ 2| $\le$ 1}: $|x - 2| \le 1 \implies -1 \le x - 2 \le 1 \implies 1 \le x \le 3$. Since $3.111...$ is not in $[1, 3]$, $k$ does not lie in this set.

  • (C) {x $\in$ R : |x $-$ 1| < 1}: $|x - 1| < 1 \implies -1 < x - 1 < 1 \implies 0 < x < 2$. Since $3.111...$ is not in $(0, 2)$, $k$ does not lie in this set.

  • (D) {x $\in$ R : |x $-$ 5| $\le$ 1}: $|x - 5| \le 1 \implies -1 \le x - 5 \le 1 \implies 4 \le x \le 6$. Since $3.111...$ is not in $[4, 6]$, $k$ does not lie in this set.

  Thus, $k = \frac{28}{9}$ belongs to the set given in option (A).

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Common Mistakes & Tips

• Distinguishing Distinct vs. Identical: Always clearly identify if the items (balls) and bins (boxes) are distinct or identical. This fundamentally changes the counting method. Here, both are distinct.
• "Exactly" vs. "At Least": The word "exactly" is crucial. It means we must ensure no more and no less than the specified quantity. For "at least", the calculation would involve summing probabilities for multiple exact counts.
• Careful with Remaining Items/Bins: After fulfilling a specific condition (like placing 3 balls in $B_3$), remember to adjust the number of remaining items and available bins for subsequent distribution steps.
• Simplifying Exponents: Be proficient with exponent rules, especially when simplifying expressions involving powers of the same base, as seen in calculating $k$.

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Summary

To find the probability, we first determined the total number of ways to distribute 9 distinct balls into 4 distinct boxes, which is $4^9$. Next, we calculated the number of favorable outcomes: selecting 3 balls for box $B_3$ ($^9C_3$ ways) and then distributing the remaining 6 balls into the other 3 boxes ($3^6$ ways). The product of these two gives the total favorable outcomes. The probability was then found by dividing favorable outcomes by total outcomes. By equating this probability to the given form $k \left( \frac{3}{4} \right)^9$, we solved for $k$, finding $k = \frac{28}{9}$. Finally, by evaluating the options, we determined that $k$ lies in the set defined by $|x - 3| < 1$.

The final answer is $\boxed{\text{A}}$