Let 9 distinct balls be distributed among 4 boxes, $\mathrm{B}_{1}, \mathrm{~B}_{2}, \mathrm{~B}_{3}$ and $\mathrm{B}_{4}$. 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) $\{x \in R:|x-2| \leq 1\}$

(B) $\{x \in R:|x-5| \leq 1\}$

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Probability

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 (\frac{3}{4})^{9}

then

k

lies in the set :

Answer: (C)

Each ball independently falls into one of the four boxes, so the number of balls landing in

B_{3}

is binomial with

n = 9

and

p = \frac{1}{4}

P = (3 9) (\frac{1}{4})^{3} (\frac{3}{4})^{6} = \frac{84 \cdot 3 ^{6}}{4 ^{9}}

Writing this in the form

k (\frac{3}{4})^{9} = k \frac{3 ^{9}}{4 ^{9}}

gives

k = \frac{84 \cdot 3 ^{6}}{3 ^{9}} = \frac{84}{27} = \frac{28}{9}

Since

\frac{28}{9} = 3.11 \dots

, the value of

k

satisfies

∣ k - 3∣ < 1

, so

k

lies in

{x \in R : ∣ x - 3∣ < 1}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option