Advertisement
Help Keep Boards Alive. Support us by going ad free today. See here: https://subscriptions.boards.ie/.
If we do not hit our goal we will be forced to close the site.

Current status: https://keepboardsalive.com/

Annual subs are best for most impact. If you are still undecided on going Ad Free - you can also donate using the Paypal Donate option. All contribution helps. Thank you.
https://www.boards.ie/group/1878-subscribers-forum

Private Group for paid up members of Boards.ie. Join the club.

Basic Probability Question

  • 24-12-2012 02:45AM
    #1
    jc2008
    Registered Users, Registered Users 2 Posts: 246 ✭✭


    This should be easy, but I'm not too sure.

    4 people choose random integers between 1 and 12 inclusive.
    (a) Probability that 3 (but not 4) people choose the same integer
    (b) Probability exactly two different are chosen.

    What I have:
    (a)
    Pr(3 pick same) = (12/12) * (1/12) * (11/12) * (1/12) = 0.00637
    Is this correct?

    (b)
    Pr(2 integers exactly) = (12/12) * (11/12) * (2/12) * (2/12) = 0.0255
    I don't think this is right though - an alternative method

    Pr(2 integers exactly) = (12 C 2)/(12 C 4) = 0.1333

    Giving me drastically different answers, neither of which are right I think. Do you know where I'm going wrong?


Comments

  • #2
    Yakuza
    Registered Users, Registered Users 2 Posts: 5,130 ✭✭✭Yakuza


    The first one can be done as a Bernoulli trial with the probabilty of success of 1/12. What are the chances of getting 3 sucesses in 4 trials?

    The second one doesn't fit that pattern, but can be done from first principles easily enough I think.
    There are 12 options for the first number, and 11 options for the second number. Lastly, how many ways can you arrange 4 things when there are two pairs of equal items?
    4!/(2! * 2!)
    If you multiply all those together and divide by 12^4 (the total number of possibilites of 4 people picking from 12 numbers) should give you the right answer.


Advertisement