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

Given:

9 balls are distributed among 4 boxes B₁, B₂, B₃, B₄.

Probability that box B₃ contains exactly 3 balls is:
k (3/4)⁹

Find the value of k.

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Step 1: Total number of possible distributions

Each of the 9 balls can go into any of the 4 boxes.

Total number of possible outcomes = 4⁹

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Step 2: Number of favourable outcomes

Exactly 3 balls must be placed in box B₃.

Number of ways to choose 3 balls from 9:

C(9, 3) = 84

The remaining 6 balls can be placed in any of the remaining 3 boxes.

Number of ways = 3⁶

Total favourable outcomes:

84 × 3⁶

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Step 3: Calculate probability

Probability =

(84 × 3⁶) / 4⁹

Given probability =

k (3/4)⁹

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Step 4: Compare both expressions

(84 × 3⁶) / 4⁹ = k (3/4)⁹

Cancelling common terms:

k = 84

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Step 5: Check given options

|x − 1| < 1 ⇒ 0 < x < 2

|x − 2| ≤ 1 ⇒ 1 ≤ x ≤ 3

|x − 3| < 1 ⇒ 2 < x < 4

|x − 5| ≤ 1 ⇒ 4 ≤ x ≤ 6

None of the given intervals contain 84.

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Final Answer:

The value of k is,
k = 84