I could do with some help with this question.
In a bag there are 18 paper notes. On five of them there is the digit 2, on seven the digit 3, and on six the digit 5. A man takes 3 notes by random. If the product of the notes is even, he wins 25 dollars. If for each game he pays 6 dollars (to participate), what is the average of profit he has after 90 games?
I am not sure how to calculate the P(Win) and how to calculate E[Profit] afterwards. I tried a tree diagram, and got 27 options, from which 19 gives a product of 19. However, when I grouped the notes to even and odd, I got a different results. Can you please assist with solving this problem?
Thank you !
Observe that the product of three numbers is odd iff all of them are odd. So, probability to lose is $\dfrac{13\choose3}{18\choose 3}$. So, the probability to win is $$1-\dfrac{13\choose3}{18\choose 3}=1-\dfrac{13\cdot12\cdot11}{18\cdot17\cdot16}=1-\dfrac{143}{408}=\dfrac{265}{408}$$. Thus, expected profit of a single game is $25\cdot\dfrac{265}{408}-6$
So, after $90$ games, the expected profit is $$90\cdot\left(25\cdot\dfrac{265}{408}-6\right)$$