Results 1 to 12 of 12

Math Help - Probabilty

  1. #1
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318

    Probabilty

    Fifteen tickets are numbered 1,2,3....,15 respectively.Seven tickets are selected at random one at a time with replacement.Find the probability that the largest number appearing on a selected ticket is 9.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie vinay's Avatar
    Joined
    Jun 2009
    From
    Hyderabad, India
    Posts
    15
    question is with or with-out replacement..


    If with replacement...

    = (1/15)(9/15)^6

    at least once 9 should come = 1/15
    after that any thing form 1-9, 6 times = (9/15)^6

    !------------------------------------


    If with out replacement

    = 9C7 / 15C7

    selecting combinations = 15C7
    selecting from 1-9 is 9C7
    Last edited by vinay; August 17th 2009 at 04:02 AM. Reason: correcting..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    The question clearly states with replacement.

    Answer is \frac{9^7-8^7}{15^7}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie vinay's Avatar
    Joined
    Jun 2009
    From
    Hyderabad, India
    Posts
    15
    ok if it is with replacements... then how come we can "Seven tickets are selected at random one at a time" at a time ?? .. thats why i got confused.....

    bye the way can you please give the explainaiton to the answer that you gave ??
    Last edited by vinay; August 14th 2009 at 05:55 AM. Reason: correctring
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    Quote Originally Posted by vinay View Post
    ok if it is with replacements... then how come we can "Seven tickets are selected at random one at a time" at a time ?? .. thats why i got confused.....

    bye the way can you please give the explainaiton to the answer that you gave ??
    It probably means that once a ticket is drawn the number on it is noted and it is put back into the bag.Then another ticket is drawn and so on the process is repeated 7 times.
    I only know the answer.It's explanation is still out of reach to me.

    I hope somebody will point out the flaw in your answer.I had done it the same way as you did
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1
    The following is the answer to the original question is:
    \sum\limits_{k = 1}^7 {\binom{7}{k}}{\left( {\frac{8}{{15}}} \right)^{7 - k} \left( {\frac{1}{{15}}} \right)^k}  =  0.057194 = \frac{{9^7  - 8^7 }}{{15^7 }}

    Now why? The summation calculates the probabilities of having from 1 to 7 nines and all the other numbers less that nine.

    But note that is the same as the other proposed answer: \frac{{9^7  - 8^7 }}{{15^7 }}.
    There are 9^7 ways to select numbers that do not exceed 9.
    There are 8^7 ways to select numbers that are less than 9.
    So there are 9^7-8^7 ways to select numbers that do not exceed 9 having at least one nine.
    There are at total of 15^7 ways to select the numbers .
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    Quote Originally Posted by vinay View Post
    question is with or with-out replacement..


    If with replacement...

    = (1/15)(8/15)^6

    at least once 9 should come = 1/15
    after that any thing form 1-7, 6 times = (8/15)^6
    What is wrong with this argument?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1
    Quote Originally Posted by pankaj View Post
    If with replacement...= (1/15)(8/15)^6
    at least once 9 should come = 1/15
    after that any thing form 1-7, 6 times = (8/15)^6
    What is wrong with this argument?
    That is the probability of getting exactly one nine in a particular position.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    Quote Originally Posted by Plato View Post
    That is the probability of getting exactly one nine in a particular position.
    So can \frac{1}{15}\left(\frac{9}{15}\right)^6 be said to be correct
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1
    Quote Originally Posted by pankaj View Post
    So can \frac{1}{15}\left(\frac{9}{15}\right)^6 be said to be correct
    Absolutely not. \frac{9^7 -8^7}{15^7} is the correct answer to the OP.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie vinay's Avatar
    Joined
    Jun 2009
    From
    Hyderabad, India
    Posts
    15
    does not give the probability of giving 9 alteast once... ????
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1
    Quote Originally Posted by vinay View Post
    does not give the probability of giving 9 alteast once... ???
    The answer is no for all the reasons listed above.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probabilty help
    Posted in the Statistics Forum
    Replies: 1
    Last Post: December 11th 2009, 12:21 AM
  2. Help for probabilty
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: October 30th 2008, 09:49 AM
  3. Probabilty that x+y+z=3
    Posted in the Statistics Forum
    Replies: 6
    Last Post: September 23rd 2008, 12:39 PM
  4. probabilty
    Posted in the Statistics Forum
    Replies: 6
    Last Post: September 15th 2008, 03:39 PM
  5. Probabilty....?
    Posted in the Math Topics Forum
    Replies: 6
    Last Post: July 9th 2005, 11:40 PM

Search Tags


/mathhelpforum @mathhelpforum