Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - Difficult (for me, that is) Combinatorics/Probability problem

  1. #1
    Junior Member
    Joined
    Sep 2010
    Posts
    36

    Difficult (for me, that is) Combinatorics/Probability problem

    Four people are going for an interview between 9am and 10am. Each person is to be interviewed in two separate rooms for a total of 30 minutes. They are interviewed for 15 minutes in room 1. And they are interviewed for 15 minutes in room 2. Each person is interviewed by themselves (no doubling up occurs in a room in a specified time slot).

    1) - How many ways can the interviews be scheduled during the one hour period.
    2) - If a person arrives at 9am, what is the probability they will have their interviews consecutively (eg 1 at 9am, the other at 9:15.... or 1 at 9:30, the other at 9:45 etc...)
    2) - If someone turns up at 9am, what is the probability they will be able to leave after the third interview since they need to attend another appointment somewhere else?



    Any help with these ones will be greatly appreciated!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Oct 2009
    Posts
    295
    Thanks
    9
    I believe the answer to 1 is:

    {{8}\choose{4,4}}=\frac{8!}{(4!)(4!)}=...

    This is because we are conducting 8 total interviews (numerator) in two rooms that have four slots each (denominator).

    2 and 3 seem like you'd use the negative binomial distribution to solve.
    Last edited by downthesun01; October 6th 2010 at 10:35 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Apr 2009
    Posts
    677
    Nopes I don't think so. There are more restrictions in the problem.
    We need to use de-arrangement
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2010
    Posts
    36
    So do you actually know how to do it? My friend and I tried to get it and we got 216.... For q1.... And 50% for the second answer. What do you think?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Apr 2009
    Posts
    677
    216 is correct
    Haven't tried Q2 Q3
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Oct 2009
    Posts
    295
    Thanks
    9
    How do you do question 1?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Apr 2009
    Posts
    677
    Room 1: Slot 1,2,3,4
    Room 2: Slot 1,2,3,4

    You can arrange 4 people to occupy Slot's in Room 1 in 4! ways
    Once done, Slots in Room 2 can give given to 4 people, with care that no one occupies the same slot in both room 1 and room 2. This just de-arrangement for n = 4 which is equal to 9.
    So 4!*9 is the ans
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Oct 2009
    Posts
    295
    Thanks
    9
    Thanks. I'm not familiar with the term "de-arrangement." I tried googling it but I'm not finding anything. Can you explain further?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Sep 2010
    Posts
    36
    Amen_cc clearly you have an idea on how to do this stuff (as opposed to me) any chance you could help a brother out and have a crack at the other questions.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Sep 2010
    Posts
    36
    Derangements are permutations of a set such that there is no element in it's original position. You need to use this in this problem because you can't have a person sitting in both 9:00 am timeslots etc. Google 'derangements' (not de arrangemnts) and you'll find a formula on wiki
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Apr 2009
    Posts
    677
    Derangement - Wikipedia, the free encyclopedia
    Plz look at this - it explains it pretty well
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Apr 2009
    Posts
    677
    sure i will look at other questions as well
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Sep 2010
    Posts
    36
    Quote Originally Posted by aman_cc View Post
    sure i will look at other questions as well
    Alright thanks heaps! Really appreciate it. Hope to hear from you soon
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    Sep 2010
    Posts
    36
    Come on peoples any help here would be MUCH appreicated
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Super Member
    Joined
    Apr 2009
    Posts
    677
    Can u plz explain Q2 and Q3 - I have not really followed them. Also it would be good if u cld share your working
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Difficult probability problem
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: August 30th 2011, 08:07 AM
  2. Probability/Combinatorics Q
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: April 10th 2010, 07:11 AM
  3. Difficult Combinatorics Problem, Help PLZ
    Posted in the Statistics Forum
    Replies: 2
    Last Post: April 11th 2008, 05:26 AM
  4. Probability/Combinatorics Problem
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: January 16th 2008, 07:54 PM
  5. Very difficult combinatorics problem
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: December 20th 2007, 03:15 AM

Search Tags


/mathhelpforum @mathhelpforum