# Permutation homework

1. Jan 9, 2008

### acme37

This is problem 4 from section 2.3 of Fundamentals of Probability by Saeed Ghahramani.

1. The problem statement, all variables and given/known data

Robert has eight guests, two of whom are Jim and John. If the guests will arrive in a random order, what is the probability that John will not arrive right after Jim?

2. Relevant equations

...

3. The attempt at a solution

There are 8! ways for the guests to arrive. There are 7 ways for John to arrive right after Jim. The probability of John arriving right after Jim is then 7/8!, and the probability of John not arriving right after Jim is 1-7/8!=0.9998.

I am pretty sure it is right, but not 99.98% sure.

2. Jan 9, 2008

### Dick

That's wrong. There aren't 7 ways for John to arrive right after Jim. There are 7 times the number of ways for all of the other guests to arrive.

3. Jan 9, 2008

### acme37

True. So,

1 - (7*6!)/8! = .875

4. Jan 9, 2008

### Dick

Much better. I'm 99.98% sure that's right.

5. Jan 17, 2012

### mv1986

Re: Permutation

There are only 7 ways for John to arrive right after Jim, apart from this detail being insufficient for the solution, or?

6. Jan 17, 2012

### Dick

Re: Permutation