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5th Combinatorial question

  1. Jul 18, 2007 #1
    1. The problem statement, all variables and given/known data
    To win Division 1 in the game of Tattslotto, the player must have the same 6 numbers (in any order) as those that are randomly drawn from the numbers 1 to 45. A Division 2 prize requires that the player’s ticket must have 5 of the 6 winning numbers and a Division 3 prize requires that the player has 4 of the 6 numbers drawn. Calculate the probability that the player’s 6 numbers will contain at least a Division 3 prize.


    2. Relevant equations
    Pr(event) = Favourable possibilities/total possibilities

    C means ‘choose’ i.e. nCr=n!/(r!(n-r)!)


    3. The attempt at a solution
    Assume player has chosen their 6 numbers and the officials have drawn the 6 numbers.

    Favourable possibilities = {6C6=1, 6C5=6, 6C4=15}
    Total possibilities = 45C6

    Pr(event) = (1+6+15)/45C6 = 22/8145060

    But the answers suggested 1135/8145060
     
  2. jcsd
  3. Jul 18, 2007 #2
    have you learned the inclusion /exclusion principle...if not i suggest looking it up.

    If you cant find it then think about what are the relations between
    6C6 6C5 6C4

    also your summation is wrong...reread the quesiton particularly the last 2 statements.
     
  4. Jul 18, 2007 #3

    Dick

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    Funny. I get 1135/814506. Your stated answer is not in lowest terms, so I suspect a typo. You are only accounting for the ways of choosing winning numbers, you have to choose the correct number of losing numbers in each case as well.
     
  5. Jul 18, 2007 #4
    I am a bit lost here. I understand the inclusion/exclusion principle but don't see how it's applied here.
     
  6. Jul 18, 2007 #5

    Dick

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    I don't either, but there are many more ways of picking four winning numbers than 6C4. You have two more numbers to choose.
     
    Last edited: Jul 18, 2007
  7. Jul 18, 2007 #6
    my bad read the question wrong.
     
  8. Jul 19, 2007 #7
    Right. To get either 1st, 2nd or 3rd prize, the total number of combinations is 6C6*39C0 + 6C5*39C1 + 6C4*39C2. Total possibilities is 45C6.

    So its (1+234+11115)/8145060 = 1135/814506
     
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