Host of party asks 6 guests to choose 1 item from a list of 10. What is probability at least 2 guests will bring same item?

It is assumed that every guest choose 1 of 10 items randomly.

First, find the probability that nobody bring the same item.

1st guest : any item is acceptable, p=1.

2nd guest : different from the item of 1st guest, p=9/10.

3rd guest : different from the items of 1st and 2nd guests, p=8/10.

......and so on.

Therefore, the probability is 1(9/10)(8/10)(7/10)(6/10)(5/10)=0.1512

The prabability at least 2 guests choose the same item is 1-0.1512=0.8488.

Use the concept of the complement of an event

P(at least 2 bring same item) = 1 - P(all bring different items)

= 1- 10/10 *9/10 *......*5/10 = 0.8488 <----

If it was a worst case, everyone could bring the same thing. HIghly unlikely but could happen.

20 percent