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!}

$$

Therefore the answer is

$$

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

$$

## 1 comment:

hi i have a question You are playing a dice game with your friend by rolling two standard dice and recording the sum of two numbers. You will win when two consecutive outcomes are 7. Your friend will win when three consecutive outcomes are in increasing order. You will continue rolling until one of you will win. What is the probability that you will win?

Examples:If the outcomes are 10,4,6,6,7,7 you will win. If the outcomes are 7,3,7,9 your friend will win.

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