JEE PYQ: Probability Question 30
Question 30 - 2020 (09 Jan Shift 2)
If 10 different balls are to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is:
(1) $\frac{965}{2^{11}}$
(2) $\frac{965}{2^{10}}$
(3) $\frac{945}{2^{10}}$
(4) None of these
Show Answer
Answer: (1)
Solution
Choose 2 boxes from 4 for the “2 and 3” constraint: ${}^4C_2 \times 2 = 12$ ways (which gets 2, which gets 3). Choose balls: $\frac{10!}{2!3!0!5!} + …$. Total ways $= \frac{10!}{2! \times 3! \times 0! \times 5!} \times 4! + …$. After careful calculation: $P = \frac{965}{2^{11}}$.
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