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In some lottery, there are $45$ numbers to enter, i.e. from $1$ to $45$, and $7$ winning numbers and $2$ supplementary numbers to draw. To win the first prize, one must have as same $7$-number in his/her ticket as the $7$-winning-number has been drawn. Order of the $7$ numbers doesn't matter. So the inverse of probability of winning the first prize is

$x_1=\dfrac{45\times 44\times 43\times 42\times 41\times 40\times 39}{7\times 6\times 5\times 4\times 3\times 2\times 1}=45379620$. Easy!

To win the second prize, one must have as same $6$ winning numbers in his/her ticket of the drawn-$7$-winning-number, plus one of the drawn-$2$-supplementary-number among his/her $7$-number-ticket. The inverse of probability of winning this prize is $x_2=3241401$. Not easy! I can't solve how this number, $x_2$, comes out. The results without step-by-step solution is here.

Please help me; Thank you very much.

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  • $\begingroup$ Your calculation of probability x_1 is inverted. Please fix. $\endgroup$ – David G. Stork Jan 25 '15 at 6:15
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The result you have for $x_1$ is a straightforward 45-choose-7 combinations.

That still applies to the ticket selection for $x_2$, but for second prize there is more variety in what they can match. There are now more options to match against - in fact ${7\choose 6}{2\choose 1}=7\times 2 = 14$ options. Take the ratio of the ticket choice against the successful options and you're home.

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To win second prize, you get to choose which of the 7 numbers you did not hit, then which of the 2 supplementary numbers you did hit. The odds are then $$ 7 \times 2 \times \frac{7\times 6 \times 5 \times 4 \times 3 \times 2 \times 1} {45\times 44 \times 43 \times 42 \times 41 \times 40 \times 39} = \frac{14}{45379620}=\frac{1}{3241401} $$

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