Question

Let 9 distinct balls be distributed among 4 boxes, B₁, B₂, B₃ and B₄. If the probability than B₃ contains exactly 3 balls is k \((\frac{3}{4})^9\) then k lies in the set:

(1) {x ∈ R : |x – 3| < 1} 

(2) {x ∈R : |x – 2| \(\le\) 1} 

(3) {x ∈R : |x – 1| < 1} 

(4) {x ∈R : |x –5| \(\le\) 1}

Answer 1

Correct option (1) {x ∈ R : |x – 3| < 1} 

required probability = \(\frac{^9C_3.3^6}{4^9}\)

= \(\frac{^9C_3}{27}\). \(\left(\frac{3}{4}\right)^9\)

= \(\frac{28}{9}.(\frac{3}{4})^9\)

\(\Rightarrow\) k = \(\frac{28}{9}\)

Which satisfies |x-3|<1