ProbabilityHardBloom L3
Question
If n different objects are distributed among ' $n + 2$ ' persons, then
Options
A.Probability that exactly 2 persons will get nothing is $\frac{(n + 1)!}{2(n + 2)^{n - 1}}$
B.Probability that exactly 2 persons will get nothing is $\frac{(n + 1)!}{(n + 2)^{n - 1}}$
C.Probability that exactly 3 persons will get nothing is $\frac{\ ^{n + 2}C_{3}\ ^{n}C_{2}(n - 1)\text{~}\text{!}\text{~}}{(n + 2)^{n}}$
D.Probability that exactly 3 persons will get nothing is $\frac{(n + 1)n^{2}(n - 1)^{2}}{12(n + 2)^{n - 1}}$
Solution
Sol. [Key Concept: Distribute $n$ objects among $n + 2$ persons; count the exactly-2-empty and exactly-3-empty cases.]
Total distributions: $(n{+ 2)}^{n}$.
Exactly 2 persons empty:
choose them $C(n + 2,2)$ ways and send the n objects one-each to the other n ($n!$):
$P = \frac{C(n + 2,2)\ n!}{(n{+ 2)}^{n}} = \frac{(n + 1)!}{2(n{+ 2)}^{n - 1}}$ — (A).
Exactly 3 empty: $C(n + 2,3)$ × surjections of n onto n−1 ($C(n,2)(n - 1)!$): $P = \frac{C(n + 2,3)\ C(n,2)(n - 1)!}{(n{+ 2)}^{n}}$ — (C), which simplifies to (D). → (A), (C), (D).
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