Question

at a party n people throw their name tags on a table. the name tags are mixed up and then each people randomly and simultaneously selects one. find the expected number of people who select their own name tag.

Answer 1

The expected number of people who will select their own name tag at a party, regardless of the number of attendees, is 1. This is derived using linearity of expectation in the context of the derangement problem in probability theory.

Step-by-step explanation:

The situation described is an example of a well-known problem in probability theory known as the derangement problem or hat-check problem. To find the expected number of people who will select their own name tag, we can use the concept of linearity of expectation.

We define an indicator variable Xi for each person i, which is 1 if person i selects their own name tag and 0 otherwise. The expected value of Xi is the probability that person i selects their own name tag, which is 1/n.

Because this holds for each person independently, the expected number of people who select their own name tag, E[X], is the sum of the expected values of the indicator variables.

E[X] = E[X1] + E[X2] + ... + E[Xn] = n*(1/n) = 1.

Therefore, regardless of the number of people at the party, the expected number of people who select their own name tag is 1.