Let's say that in a lottery, 4 balls are drawn at random from a pot of 10 balls, numbered from 0 to 9. The drawn balls are not put back in the pot. If your ticket matches the 4 drawn balls you win the jackpot. However if the last number on your ticket ends with one of the four drawn balls, you get your money back. My question is if you have 2 tickets and both ending in different numbers, what are the chances that:
I get my money back for exactly one ticket,
I get my money back for at least one of the tickets,
I get my money back for both of the tickets
Thanks
Best Answer
For $i=1,2$ let $E_{i}$ be the event that you get your money back on the $i$-th ticket (i.e. the last number of $i$-th ticket agrees with one of the $4$ distinct numbers drawn). Then:
$P\left(E_{1}\wedge E_{2}\right)=P\left(E_{2}\mid E_{1}\right)P\left(E_{1}\right)$.
Here $P\left(E_{1}\right)=\dfrac{4}{10}$ and $P\left(E_{2}\mid E_{1}\right)=\dfrac{3}{9}$ (i.e the probability that - under condition that my first the last number on first ticket belongs to numbers drawn - this is also the case for the last number on second ticket).
This leads to $P\left(E_{1}\wedge E_{2}\right)=\dfrac{12}{90}=\dfrac{2}{15}$.
Likewise $P\left(E_{1}\wedge E_{2}^{c}\right)=P\left(E_{2}^{c}\mid E_{1}\right)P\left(E_{1}\right)=\dfrac{6}{9}\times\dfrac{4}{10}=\dfrac{4}{15}$.
Consequently the probability of money back exactly once will equal $P\left(E_{1}\wedge E_{2}^{c}\right)+P\left(E_{1}^{c}\wedge E_{2}\right)=\dfrac{8}{15}$.
Finally the probality of no money back at all is: $1-\dfrac{2}{15}-\dfrac{8}{15}=\dfrac{1}{3}$.