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.
Best Answer
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: $$\begin{align*}P(\text{Win 1st or 2nd}|\text{Won 1st or 2nd last week}) & = \dfrac{P(\text{Win 1st or 2nd} \cap \text{Won 1st or 2nd last week})}{P(\text{Won 1st or 2nd last week})} \\ & = \dfrac{P(\text{Win 1st or 2nd})P(\text{Won 1st or 2nd last week})}{P(\text{Won 1st or 2nd last week})} \\ & = P(\text{Win 1st or 2nd}) \\ & = \dfrac{1}{_{10}P_3} + \left( \dfrac{1}{_{10}C_3} - \dfrac{1}{_{10}P_3} \right) \\ & = \dfrac{1}{_{10}C_3} = \dfrac{1}{120} \end{align*}$$