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Below is a problem I think that I have solved correctly, but cannot seem to get the correct answer. Any help would be greatly appreciated.

You pay $\$13.00$ for a ticket. When you buy a ticket, seven numbers are drawn at random without replacement from the set $\{1,2,3,4,\dots, 46,47\}$. If $x$ is the number of prime numbers selected, then you win nothing if $x = 0$, $x = 1$, $x = 4$, or $x= 6$. You get back $\$1.00$ if $x = 2$, $\$10$ if $x = 3$, $\$100$ if $x = 5$, and $\$10,000$ if $x = 7$. Your expected loss from a single ticket is:

My proposed solution: I can only win a sum of money when I have $2$, $3$, $5$, $7$ primes in my list. Thus, the chances of getting these would be:

$$P(2\text{ primes}) = \frac{{33 \choose 5}{14 \choose 2}}{{47 \choose 7}} = 0.34341$$

$$P(3\text{ primes}) = \frac{{33 \choose 4}{14 \choose 3}}{{47 \choose 7}} = 0.23683$$

$$P(5\text{ primes}) = \frac{{33 \choose 2}{14 \choose 5}}{{47 \choose 7}} = 0.01680$$

$$P(7\text{ primes}) = \frac{{33 \choose 0}{14 \choose 7}}{{47 \choose 7}} = 0.00005457$$

With $2$ primes, we win $\$1$ leading to a profit of $\$1\times0.34341 = \$0.34341$

With $3$ primes, we win $\$10$ leading to a profit of $\$10\times0.23683 = \$2.3683$

With $5$ primes, we win $\$100$ leading to a profit of $\$100\times0.01680 = \$1.68076$

With $7$ primes, we win $\$10000$ leading to a profit of $\$10000\times0.00005457 = \$0.5457$

Thus our expected value to win should be $-\$13$ (the cost of the ticket) $+$ the above amounts, which yields $-\$8.06$.

I'm sure there is some sort of error in my reasoning, but I cannot figure out where it is.

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  • $\begingroup$ Your expected loss is $8.06 $\endgroup$ – The Chaz 2.0 Jan 22 '15 at 22:10
  • $\begingroup$ There are $15$ primes in the set. They are $$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$$ Thus, you should be making selections from $32$ non-primes and $15$ primes. $\endgroup$ – N. F. Taussig Jan 22 '15 at 22:36
  • $\begingroup$ @N.F.Taussig You are completely right! Somehow stupid me skipped the prime $7$. I think you should post this as an answer. $\endgroup$ – Peter Woolfitt Jan 22 '15 at 22:43
  • $\begingroup$ Yes, I completely missed 43 in my list. You are completely right, and that solved my issue!! A loss of $6.67 is correct. Thank you and Peter for your help!!! $\endgroup$ – user160298 Jan 22 '15 at 22:50
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The strategy you applied is sound. However, the set $\{1, 2, 3, \ldots, 47\}$ contains $15$ primes. They are $$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$$ Therefore, you should have been making selections from $15$ primes and $32$ non-primes.

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