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    ccny_miah's Avatar
    ccny_miah Posts: 15, Reputation: 1
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    #21

    Jul 31, 2012, 11:40 AM
    I don't know where to start.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #22

    Jul 31, 2012, 11:47 AM
    The numerator is the number of ways that you can select the winning 5 numbers times the number of ways you can select the correct extra number. The denominator is the total number of possible ways that 5 balls can be drawn from 53, times the number of ways that the one additional ball can be drawn from 42.
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    ccny_miah Posts: 15, Reputation: 1
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    #23

    Jul 31, 2012, 11:59 AM
    Quote Originally Posted by ebaines View Post
    The numerator is the number fo ways that you can select the winning 5 numbers times the number if ways you can select the correct extra number. The denominator is the total number of possible ways that 5 balls can be drawn from 53, times the number of ways that the one additiobnal ball can be drawn from 42.
    Can you please show me... I really confused on this.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #24

    Jul 31, 2012, 12:44 PM
    I'll give you a different example that illustrates the thought process. What's the probability of dealing a royal straight spade flush from a deck of 52 cards? A royal straight spade flush consists of the ace of spades, king of spades, queen of spades, jack of spades, and 10 of spades. It doesn't matter what order the five cards are dealt in. The number of ways those five cards can be dealt is 5! The total number of possible ways to deal five cards is 52 x 51 x 50 x 49 x 48 = 52P5 (permutations of 52 cards 5 at a time). Hence the odds of dealing a royal straight flush is 5!/(52P5) = 0.0000385%.
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    ccny_miah Posts: 15, Reputation: 1
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    #25

    Aug 1, 2012, 06:35 AM
    Quote Originally Posted by ebaines View Post
    The numerator is the number of ways that you can select the winning 5 numbers times the number of ways you can select the correct extra number. The denominator is the total number of possible ways that 5 balls can be drawn from 53, times the number of ways that the one additional ball can be drawn from 42.
    53C5 x 42C1

    53!x42/(53-5)!(5!)

    (53x52x51x50x49)(42)/5x4x3x2x1
    = 120,526,770?

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