IAI November 2010 Examination Q3

Discussion in 'CT3' started by paryas.bhatia, May 8, 2013.

  1. paryas.bhatia

    paryas.bhatia Member

    Can we use combinations for solving this problem? If yes, please mention the solution.
     
  2. asmkdas

    asmkdas Member

    (a)
    Since p = 1/365 and q = 364/365
    The probability of occurring one person's birthday is [1/365]
    The probability of occurring two person's birthday is [1/365] * [1/365]
    In a complete year
    the probability of occurring two person's birthday is
    [ [1/365] *
    [1/365] * 365 ] = 0.002740

    (b)
    Say X is the number of birthdays, and hence X ~ Bin (3 , 1/365)
    In a complete year occurring of at least two person birthdays is
    P[ X => 2] * 365 = [ P [ X = 2 ] +
    P [ X = 3] ] * 365
    = [3C2 * (1/365)^2 * (364/365)^(3 - 2) + 3C3
    * (1/365)^3 * (364/365)^0 ]
    = 0.008204

    Similarly

    Say X is the number of birthdays, and hence X ~ Bin (4 , 1/365)
    In a complete year occurring of at least two person birthdays is
    P[ X => 2] * 365 = [ P [ X = 2 ] +
    P [ X = 3] + P [ X = 4] ] * 365
    = [4C2 * (1/365)^2 * (364/365)^(4 - 2) + 4C3
    * (1/365)^3 * (364/365)^1 + 4C4 * (1/365)^4 * (364/365)^0 ] * 365
    = 0.016356

    (c)
    Say X is the number of birthdays, and hence X ~ Bin (15 , 1/365)
    In a complete year occurring of at least two person birthdays is
    P[ X => 2] * 365 = [ 1 - P[ X < 2] ] * 365 = 0.25
     
  3. paryas.bhatia

    paryas.bhatia Member

    Why are we multiplying each probability with 365?
     
  4. asmkdas

    asmkdas Member

    Actually in the question it was asked us to get the probability for 365 days.
     
  5. paryas.bhatia

    paryas.bhatia Member

    Sorry but I didn't get you :confused:
     
  6. asmkdas

    asmkdas Member

    In the question it was asked us to get the probability as a whole i.e., for 365 days. That's why we have multiplied it with 365 days.
     

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