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A weekly lottery consists of 3 numbers drawn from the digits 0 through 9 with no repetition of digits. The first prize goes to the person with the correct sequence. Second prizes go to people with the correct digits in some other sequence. You buy a ticket.

a) What is the probability of winning the first prize?

b) What is the probability of winning the second prize?

c) If you win a first or second prize last week, what is your probability of winning a first or second prize this week?

I know that the odds of winning are 9!/(9-3)!=504, 1/504 but besides that, I am lost. Any help would be greatly appreciated.

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    $\begingroup$ How many possible tickets are there? How many win first prize? How many win second prize? Your $\frac 1{504}$ is not correct. It would be correct if the digits were $1$ through $9$. $\endgroup$ – Ross Millikan Feb 14 at 20:41
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    $\begingroup$ Nothing in the phrasing of the rules suggests that winning last week should have anything at all to do with winning this week. $\endgroup$ – lulu Feb 14 at 20:45
  • $\begingroup$ There are $3!=6$ ways of arranging $3$ distinct digits $\endgroup$ – Henry Feb 14 at 21:05
  • $\begingroup$ Question (c) seems to imply that the probabilities of winning each week aren't independent. Is there a missing piece of information like the player not reusing the same sequence? $\endgroup$ – Jam Feb 14 at 21:48
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a) $$\dfrac{1}{_{10}P_3} = \dfrac{(10-3)!}{10!} = \dfrac{1}{720}$$

b) $$\dfrac{3!-1}{_{10}P_3} = \dfrac{(3!-1)(10-3)!}{10!} = \dfrac{1}{_{10}C_3} - \dfrac{1}{_{10}P_3} = \dfrac{5}{720}$$

c) Because winning in the past is independent of winning in the future: $$\begin{align*}P(\text{Win 1st or 2nd}|\text{Won 1st or 2nd last week}) & = \dfrac{P(\text{Win 1st or 2nd} \cap \text{Won 1st or 2nd last week})}{P(\text{Won 1st or 2nd last week})} \\ & = \dfrac{P(\text{Win 1st or 2nd})P(\text{Won 1st or 2nd last week})}{P(\text{Won 1st or 2nd last week})} \\ & = P(\text{Win 1st or 2nd}) \\ & = \dfrac{1}{_{10}P_3} + \left( \dfrac{1}{_{10}C_3} - \dfrac{1}{_{10}P_3} \right) \\ & = \dfrac{1}{_{10}C_3} = \dfrac{1}{120} \end{align*}$$

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