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

JEE · Math · previous-year question

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

Answer

A. {x $$\in$$ R : |x $$-$$ 3| < 1}

Explanation

Required probability = $${{{}^9{C_3}{{.3}^6}} \over {{4^9}}}$$ $$ = {{{}^9{C_3}} \over {27}}.{\left( {{3 \over 4}} \right)^9}$$ $$ = {{28} \over 9}.{\left( {{3 \over 4}} \right)^9} \Rightarrow k = {{28} \over 9}$$ Which satisfies $$\left| {x - 3} \right| < 1$$

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