Question

Let 9 distinct balls be distributed among 4 boxes, $B_1, B_2, B_3$ and $B_4$. If the probability that $B_3$ contains exactly 3 balls is $k \left(\frac{3}{4}\right)^9$ then $k$ lies in the set :

• A. $\{x \in \mathbb{R} : |x - 1|<1\}$
• B. $\{x \in \mathbb{R} : |x - 2| \le 1\}$
• C. $\{x \in \mathbb{R} : |x - 3|<1\}$
• D. $\{x \in \mathbb{R} : |x - 5| \le 1\}$

Hint:

The probability of box $j$ having exactly $r$ balls in a distribution of $n$ distinct balls into $m$ boxes is $P = \binom{n}{r} (\frac{1}{m})^r (1 - \frac{1}{m})^{n-r}$. Here $n=9, m=4, r=3$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a problem of distributing distinct objects into distinct containers.
Each ball has 4 choices (boxes).
The probability of a specific box containing a certain number of balls follows the binomial distribution, where "success" is defined as a ball falling into box \(B_3\).
Step 2: Key Formula or Approach:
1. Total number of ways to distribute \(n\) balls into \(m\) boxes is \(m^n\).
2. Number of ways for \(B_3\) to have exactly \(r\) balls: Choose \(r\) balls for \(B_3\) and distribute the remaining \(n-r\) balls into the other \(m-1\) boxes.
Ways $= \binom{n}{r} (m-1)^{n-r}$.
Step 3: Detailed Explanation:
Total ways to distribute 9 distinct balls into 4 boxes = \(4^9\).
Favorable ways (exactly 3 balls in \(B_3\)):
1. Select 3 balls out of 9 for \(B_3\): \(\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\).
2. Distribute the remaining \(9 - 3 = 6\) balls into the other 3 boxes (\(B_1, B_2, B_4\)): Each of these 6 balls has 3 choices. Ways = \(3^6\).
Total favorable ways = \(84 \times 3^6\).
Probability \(P = \frac{84 \times 3^6}{4^9}\).
We are given \(P = k \left( \frac{3}{4} \right)^9\). Equating the two:
\[ k \frac{3^9}{4^9} = \frac{84 \times 3^6}{4^9} \]
Cancel \(4^9\) from both sides:
\[ k \cdot 3^9 = 84 \cdot 3^6 \Rightarrow k = \frac{84}{3^3} = \frac{84}{27} \]
Simplify by dividing by 3:
\[ k = \frac{28}{9} \approx 3.111 \]
Now check the options:
(A) \(|3.111 - 1| = 2.111>1\). False.
(B) \(|3.111 - 2| = 1.111>1\). False.
(C) \(|3.111 - 3| = 0.111<1\). True.
(D) \(|3.111 - 5| = 1.889>1\). False.
Step 4: Final Answer:
The value \(k = 28/9\) lies in the set \(\{x \in \mathbb{R} : |x - 3|<1\}\).