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 :
Solution
Required probability = ${{{}^9{C_3}{{.3}^6}} \over {{4^9}}}$<br><br>$= {{{}^9{C_3}} \over {27}}.{\left( {{3 \over 4}} \right)^9}$<br><br>$= {{28} \over 9}.{\left( {{3 \over 4}} \right)^9} \Rightarrow k = {{28} \over 9}$<br><br>Which satisfies $\left| {x - 3} \right| < 1$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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