Hello there. To solve this question, we'll have to remember some properties about probabilities.
First, remember that there are 4 suits in a standard card deck, each containing 13 different cards.
Most specifically, there are only one of each number in each suit, so this means that there are only 4 cards with the number 5 in a 52 card deck.
The probability of taking three fives in succession from this deck, with replacement, is given by the ratio between the favorable events of taking a 5 (4) and the total number of cards (52), cubed:
[tex]P(\text{taking a 5 three times in succession)}=\frac{4}{52}\cdot\frac{4}{52}\cdot\frac{4}{52}=\frac{1}{13^3}=\frac{1}{2197}[/tex]Notice the probabilities are the same for each succession because we were replacing the cards. If we weren't replacing, then the probabilites would go down as 4/52 * 3/51 * 2/50, which is not the case.
Obviously, this is the probability of winning and if you get three fives in succesion, you get $70.
To find the probability of not getting 5's three times in succession, simply subtract the probability we found from 1:
[tex]P(\text{not get 5's three times in succession)}=1-\frac{1}{2197}=\frac{2196}{2197}[/tex]Now, we multiply the probabilities by the number in dollars we would get by winning and the number in dollars (negative) we would pay for losing, respectively:
[tex]\begin{gathered} 70\cdot\frac{1}{2197}-10\cdot\frac{2196}{2197} \\ \\ \frac{70}{2197}-\frac{21960}{2197} \\ \\ -\frac{21890}{2197} \end{gathered}[/tex]By calculating the fraction, we get:
[tex]-9.9635[/tex]