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I would like to find the probability of guessing 5 out of 7 numbers. There is 34C7 possibles. The correct answer is 0.0014. For guessing 6 out 7 was easier. That was $\frac{28*6}{34C7}$. Since I don't get the other I don't understand the first one either. I think I just got that one by luck.

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I solved it myself using Hypergeometric distribution:

$$\frac{{27 \choose 2} \times {7 \choose 5}}{34 \choose 7} = 0.0014$$

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+1. Note that this is really just the multiplication principle for counting at work. Of the original 34 numbers, you want to guess 5 of a certain 7 numbers, and there are ${7 \choose 5}$ ways to do this, and for each of the ${7 \choose 5}$ ways to pick the numbers, there are ${27 \choose 2}$ ways to pick the remaining $2$ numbers. You then divide by the total number of choices which is ${34 \choose 7}$. The reason I made this comment is that referring to this as a hypergeometric situation seems to make this more complicated than it really is. –  JavaMan Nov 20 '12 at 9:45
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Consider the lottery game where there are 34 numbers and a draw of 7 numbers will take place. you aim for a 3rd prize (5 out of 7). What is the probability of winning?

Solution:

There are 34 numbers to choose from. The drawer will randomly choose 7 numbers, while you also pick out 7 numbers for your guess. You want the 3rd prize (for some reason you don't want to be first. Lol) and would want to match exactly 5 out of the 7 numbers drawn by the contest people. Note that there are 7 numbers out of the 34 numbers in our pool of numbers to choose from that would match their 7 numbers, we would consider these numbers as our "success" numbers. And there are 34 - 7 = 27 "fail" numbers.

We want to take 5 numbers out of our "success" pool. This gives us a total of $7\choose5$ ways. And since we draw 7 numbers, we still have to choose 2 from our "fail" pool. This gives us $27\choose2$. For probability of

$\displaystyle\frac{{27 \choose 2} \times {7 \choose 5}}{34 \choose 7}$

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