Serving Slices of Pizza Can someone please double check my work? Thanks!
There are 15 slices of pizza: 5 cheese, 5 cheese with pepperoni, and 5 no cheese.
Suppose the pizzas are served in a certain order from 1 to 15.
(1) What is the probability that all cheese with pepperoni slices are served within the first 10?
My work: I first choose 5 spots out of the first 10 to put them in. They can be ordered in 5! ways. The remaining 10 spots can be ordered in 10! ways. Total number of ways is 15! Answer is 8.39%.
(2) What is the probability that after serving 10 slices, only two types remain?
Kinda stuck on this one, but my thinking is the following:
Choose one type which can be done 3 ways. Place 5 of that type in the first 10. Those can be ordered 5! ways. Choose two spots in the final five which can be done 10 ways and those can be arranged in 2! ways. Arrange the remaining in 8! ways. Answer is 5.6%.
(3) What is the probability that two of each type are among the first six to be serviced?
Place all six in the first six slots. Order them in 6! ways. Order the rest in 9! ways. Answer is 0.02%.
 A: On question 2, if the interpretation is that at most two types remain after $10$ slices are served then
For one type of pizza, the probability that it finishes in first $10$ serving is $\frac{12}{143}$ as you found in question $1$. But it also includes probability that two types would have finished in first $10$ which is
$\displaystyle  \frac {10 \choose 10}{15 \choose 10} = \frac{1}{3003} $ for any two. So applying inclusion-exclusion,
$\small P(A \cup B \cup C) = P(A) + P(B) + P(C) - P (A \cap B) - P (B \cap C) - P(A \cap C) + P(A \cap B \cap C) \ $
where $A, B, C$ are events of a type of pizza finishing in first $10$ serving.
So, $\small P(A \cup B \cup C) =  3 \times \frac{12}{143} - \frac{3}{3003}$
If the question is about exactly two types to remain in last $5$, then we have all of one type served in first $10$ and for other two types, number of slices serves in first $10$ are $(4, 1), (3,2), (2,3), (1,4)$.
So, Probability $ \displaystyle = \frac{ {3 \choose 1} \times 2 \times \big({5 \choose 4}  {5 \choose 1} + {5 \choose 3} {5 \choose 2}\big)} {15 \choose 10} = \frac{250}{1001}$
On question 1, while your working is correct, you can simply look at it as
$\displaystyle  = \frac{10 \choose 5}{15 \choose 10} = \frac{12}{143}$
which is in the first $10$, we choose $5$ out of $5$ cheese with pepperoni pizza slices and $5$ out of remaining $10$ slices vs. any $10$ of $15$ slices.
For question 3, think number of ways of choosing $2$ out of $5$ pizza slices of each type vs. any $6$ slices out of $15$.
