As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?
I tried to solve this by doing (1/4)(1/3)(1/2)(1), but that didn't work. Can someone explain why that is incorrect? My thought process was that there is a 1/4 chance the first person will pick a certain number, then since there are only 3 numbers left, there is a 1/3 chance, etc... Now I'm all confused. Help?
1/4 chance the first person will pick a certain number
you dont have to find probability of first person "picking a number". The question asks about probability of picking DIFFERENT (than the rest 3) numbers. Now, lets solve your ways:
- chance for first person to pick a DIFFERENT number is: 1. As he is first person, his number is always different.
- chance for second person to chose DIFFERENT number = 3/4
- thirs: 2/4
- fourth : 1/4
p = 1 * 3/4 * 1/2 * 1/4 = 3/32 = .0937 or ~9 %