Tuesday, December 21, 2010

There are 20 person in a Christmas party. What is the probability that at least one of them get his own gift?

Solution:
Let $$A_i$$=person i get his own gift
Then we want $$P(A_1 \cup A_2 \cup \cdots \cup A_n)$$,
which is equal to
$$\sum P(A_i)-\sum P(A_i \cap A_j) + \sum P(A_i \cap A_j \cap A_k) - \cdots + (-1)^n \sum P(A_1 \cap A_2 \cap \cdots \cap A_n)$$

The general term is
$$C(n, k)\cdot \frac{(n-k)!}{n!} = \frac{1}{k!}$$

$$1-\frac{1}{2!}+\frac{1}{3!}-\cdots+\frac{1}{20!}$$