JEE Main 2021 · 25 July · morning · Single correct · +4 / −1
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( 3/4 )⁹ then k lies in the set :
- (a) x ∈ R : |x - 3| < 1
- (b) x ∈ R : |x - 2| ≤ 1
- (c) x ∈ R : |x - 1| < 1
- (d) x ∈ R : |x - 5| ≤ 1
Answer and solution
Answer (a) x ∈ R : |x - 3| < 1
Solution
Maths sir · 4 steps
Each ball picks a box on its own, so count 4 choices per ball, then fix which 3 land in B₃.
- Total outcomes. Each of the 9 distinct balls can go into any of the 4 boxes, independently. So there are 4⁹ equally likely distributions.
- Favourable outcomes. Choose which 3 balls go into B₃ in C(9, 3) = 84 ways. Each of the other 6 balls must avoid B₃, so it has 3 choices, giving 3⁶ ways.
P = (84 × 3⁶)/4⁹ = k(3/4)⁹ = (k × 3⁹)/4⁹
- Find k. Compare the numerators: k × 3⁹ = 84 × 3⁶, so k = 84/27 = 28/9 ≈ 3.11.
- So. |28/9 - 3| = 1/9 < 1, so k lies in {x : |x - 3| < 1}, which is option (a).
The idea to keep. For distinct objects into boxes, every object has its own choice of box. Pick the objects for the special box with C(n, r), and the rest avoid that box.
Answer from the NTA final answer key. Solution by PrepWiser.