A newly-opened restaurant has 5 menu items. If the first 4 customers each choose one menu item uniformly at random, the probability that the 4th customer orders a previously unordered item is a/b, where a and b are relatively prime positive integers. What is a+b?
4 answers
189
I need a-b
189
let A B C D E be the items
if 4th customer chose A then the other 3 customers can choose other items
B or C or D or E in 4^3 ways
similarly if 4th customer chose B then the others can chose in 4^3 ways
hence the total number of ways in which 4th customer chose one item and the others choose the remaining item =5* 4^3
total ways of choosing 5 items by 4 customers =5^4
therefore the required probability=(5*4^3)/5^4=64/125 .Hence m+n=189
if 4th customer chose A then the other 3 customers can choose other items
B or C or D or E in 4^3 ways
similarly if 4th customer chose B then the others can chose in 4^3 ways
hence the total number of ways in which 4th customer chose one item and the others choose the remaining item =5* 4^3
total ways of choosing 5 items by 4 customers =5^4
therefore the required probability=(5*4^3)/5^4=64/125 .Hence m+n=189
1 answer