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I have the following question.

A state lottery is called ”6/49” for the reason that players have to choose six different numbers from the numbers from 1 to 49, inclusive. One wins the first prize if he/she gets all the numbers correct, the second prize if only 5 of the 6 numbers are correct and the third price if only 4 of the 6 numbers are correct. How many times is one more likely to win the second prize than the first?

I know there are C(49,6) ticket possibilities, which is 13,983,816. My question is how do you solve the last question.

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  • $\begingroup$ en.wikipedia.org/wiki/Hypergeometric_distribution $\endgroup$ Commented May 5, 2020 at 2:46
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    $\begingroup$ @AnginaSeng could you elaborate how the hyper-geometric will work here? If the universe/‘denominator’ is 49c6 what would you put in the ‘numerator’ 48c5*1c1? I don’t think I understand how the usage of hyper- geometric would work here and appreciate your direction. Thanks $\endgroup$
    – adhg
    Commented May 6, 2020 at 1:50

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There are $6$ ways to choose which $5$ of the winning number you have picked, For each of those choices there are $43$ ways to pick the sixth number on the card, since it can be any of the number that doesn't come up. This give $6\cdot43=258$. So, $258$ times more likely, since there's only one way to choose the winning ticket.

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