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In a lottery, an urn contains 40 balls that are numbered 1, 2, ..., 40. Each week, 5 balls are drawn from the urn without replacement. To enter, one chooses 5 numbers. Anyone who correctly predicts exactly (and only) four of the five numbers (order doesn't matter) wins the jackpot. What's the probability of winning this lottery?

My approach was to do $\frac{{5 \choose 4} \times 34}{{40 \choose 5}}$, as I have to choose 4 out of the 5 balls, and there are $40-5-1=34$ ways to choose the last number so it does not match the final ball. As well, there should be $40 \choose 5$ ways to select five balls from the urn. However, this doesn't seem to work... Where have I gone wrong?

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  • $\begingroup$ There are $35$ numbers that do will not match the drawn balls , not 34. $\dfrac{{5\choose 4}{35\choose 1}}{40\choose 5}$ $\endgroup$ Commented Sep 28, 2014 at 23:41

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To win you must select exactly 4 of the 5 'matching' numbers, and 1 of the 35 'nonmatching' numbers. Count the ways to do this and divide by the count of ways to select any 5 of the 40 numbers.

$$\frac{{5\choose 4}{35\choose 1}}{40\choose 5}$$

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  • $\begingroup$ Thanks! I didn't realize that $35 \choose 1$ was the correct way to choose the final number. $\endgroup$ Commented Sep 28, 2014 at 23:53
  • $\begingroup$ Note that if you didn't put that $-1$ in your calculations, yours were right as well (${35\choose 1} = 35$). $\endgroup$
    – AlexR
    Commented Sep 29, 2014 at 0:06
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If the question is exactly 4, than I'd say out of ${40 \choose 5}$, you have 5 chances for the ball that can go wrong, and that can have 35 values. So I'd say 175 / ${40 \choose 5}$.

Admittedly, I'm not sure if I can also take reordering into account, but I don't think so.

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