Probability

Question

Katrina, Kim, Sarah, Larry, Maria, Ian, Jim, and Sergio have all been invited to a dinner party. They arrive randomly and each person arrives at a different time.

a. In how many ways can they arrive?

b. In how many ways can Katrina arrive first and Sergio last?

c. Find the probability that Katrina will arrive first and Sergio last?

 

Please show work.

Answer 1

a) Since there are 8 people and they arrive randomly each at a different time, then it would be 8! = 8*7*6*5*4*3*2*1 = 40320.

 

b) Now you want to know how many ways Katrina can arrive first and Sergio last. Since Katrina will always be first and Sergio will always be last, then if you have 8 slots ( because there are 8 people), Katrina will take the first and Sergio the last as follows (Assuming K stands for Katrina and S for Sergio):

 

_K_ | ___ | ___ | ___ | ___ | ___ | ___ | _S_

 

This leaves 6 empty slots (8-2=6).

Then there will be 6! ways where this is possible. Therefore the answer is 6! = 720

 

c) Now you know there are 8! ways for everyone to arrive, and 6! ways for Katrina to be first and Sergio last, then to find the probability of Katrina actually being first and Sergio last you would simply find the ratio:

 

6!/8! = 720/40320 = 0.017857 = 1/56