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mathsux

Member

Registered: 2007-08-06

Posts: 10

Probability using Counting Techniques

When 4 guys and 4 girls are arranged in a row, find the probability that:

2 guys have three seats between them?

-ms

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Probability using Counting Techniques

If Morgan and Tom have 3 seats between them, you have

M [3 kids] T [remaining 3 kids]
T [3 kids] M [remaining 3 kids]

You can arrange the 6 kids in 6! ways. Multiply that by 2 since M and T can be flipped. Then multiply all of that by 4, because as long as M and T have 3 seats between them they can occupy 4 different seats.
i.e.
T _ _ _ M _ _ _
_ T_ _ _ M _ _
_ _ T _ _ _ M _
_ _ _ T _ _ _ M

So you have

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: Probability using Counting Techniques

I think the answer is a bit more complicated than that.  Your solution only takes into account when Morgan and Tom are 3 seats apart.    What about when Bob and Fred are 3 seats apart?  Or Morgan or Fred?   There are 6 combinations of 2 boys which could be seated in the required arrangement. 

But we just can't multiply the your answer by 6 because there will be duplicates when 2 differents sets of boys have 3 seats between them.   Such as...   M B _ _ T F _ _.   

It may be easier to calculate the number of arrangements were no 2 boys are three seats apart.  Let's have the boys pick their seats.   Boy 1 can pick any of 8 seats.   Let's say he picks seat 3.   Boy 2 can then pick any of 6 seats because he can't pick seat 3 (boy 1 is already there) or seat 7 (he would be 3 seats from boy 1).  Likewise, boy 3 could pick from 4 seats and boy 4 could pick from 2 seats.   That's 8 * 6 * 4 * 2 = 384.   The 4 girls could be arranged in 4! =24 ways.   384 * 24 =9216.   And there are 8!=40320 total ways the kids would be arranged so 9216 / 40320 ~= 22.9%.  That's the chances of no 2 boys being 3 seats apart.   Subtract that from 1 to get a 77.1% chance to get the odds of 2 boys sitting 3 seats apart.   

That sounds a little high to me.   Maybe not though because if we multiplied Identity's answer by 6 we would get around 86%.   Can someone else confirm or point out a mistake in my "solution"?

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Probability using Counting Techniques

Sorry, I forgot to mention he's a friend of mine. I have a pretty fair idea of which question he meant to ask about in the book.

But well... let's see if I can get your answer.

B1 [group of 3 girls] B2 [3 remaining kids]

4 * [4 * 3 * 2] * 3 * [1 * 2 * 1]

Then multiply that by 4 because of movement down the row of seats of the boys.

So I'm getting

Last edited by Identity (2007-10-11 08:59:46)

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: Probability using Counting Techniques

I don't really understand your most recent post.   If the original question is asking for the odds of 2 specific boys sitting 3 seats apart, then your answer is correct.     If the question is what are the odds of any 2 of the 4 boys sitting 3 seats apart, then I think my answer is correct.

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: Probability using Counting Techniques

I  think I see what you were doing in post #4 now.   It looks like you are adding in the requirement that 3 girls are sitting in the the 3 seats between the 2 guys.  That wasn't stated in the original question but if you have the same books as mathsux, maybe that is what he/she meant.   

I used a little brute force to double-check my answer as the question was originally stated.   Consider the number ways you can arrange the letters BBBBGGGG with B standing for boy and G for girl.   That would be 70 (8! / (4! * 4!).    Using Excel, I generated those 70 possible combinations.  I then counted how many of the seating arrangements had 2 boys sitting 3 seats apart.   54 of the 70 (=~ 77.1%) met the requirements.