7) I'm assuming any pizza comes with cheese by default and that it shouldn't technically count as a topping on its own. It's probably safe to also assume that a pizza with sausage and pepperoni, for instance, is the same as a pizza with pepperoni and sausage, i.e. the order of toppings does not matter.
Then the number of pizzas that can be made with no more than 5 toppings would be
[tex]\displaystyle\sum_{t=0}^5\frac{12!}{t!(12-t)!}=1586[/tex]
8) Let [tex]B[/tex] be the random variable representing the earning/loss after drawing a bead. [tex]B[/tex] has the probability mass function
[tex]p(B=b)=\begin{cases}\dfrac3{12}=\dfrac14&\text{for }b=3\\\\\dfrac7{12}&\text{for }b=-1\\\\\dfrac2{12}=\dfrac16&\text{for }b=10\\\\0&\text{otherwise}\end{cases}[/tex]
The expected value for any given draw is
[tex]E[B]=\displaystyle\sum_{b\in\{3,-1,10\}}b\cdot p(B=b)=\dfrac34-\dfrac7{12}+\dfrac{10}6=\dfrac{11}6[/tex]
The game would be fair if the expectation was $0, but this one is actually in favor of the player.
