Permutation

[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.

[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.

[acme37]

True. So,

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

[Dick]

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

[mv1986]

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

[Dick]

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

So you didn't read the rest of the thread? Please do it.

[Dick]

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

I'm not sure what your question is. acme37 thought that of the 8! orders that the guests could arrive there were 7 in which John arrives right after Jim. That's wrong. The right answer is perfectly sufficient for a solution.