2

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%.

trevormil
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    Sanity check: Your answer to the second question must be at least as large as the first answer since the first answer is one of three possible cases for the second question. Also, the second questions that there must be at least one type of slice for which all five pieces appear among the first ten slices. It does not require that two types of pizza must appear in the last five slices. – N. F. Taussig Jan 31 '21 at 21:24

1 Answers1

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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 $15$.

Math Lover
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    @trevormil I agree with Math Lover's math. For question 1, another perspective is $$\frac{\binom{10}{5}}{\binom{15}{5}}.$$ The numerator represents how many ways of selecting 5 slots out of 10 for the cheese+pepperoni and the denominator represents how many ways of selecting 5 slots out of 15 for the cheese+pepperoni. Obviously, the math is identical to Math Lover's answer. ...See next comment. – user2661923 Jan 31 '21 at 23:05
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    The real issue with question 1 is : how should you set the denominator? Should you look at combinations or permutations. This is a very tricky issue that depends not only on the problem, but also on the sophistication of the problem-solver. In general, I recommend looking for an approach that you are very comfortable grasping, and where you have no problem specifying the numerator in a manner consistent with how you specify the denominator. – user2661923 Jan 31 '21 at 23:08
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    The second question asks: What is the probability that after serving $10$ slices, only two types remain? It does not say that two types must remain, just that all $5$ slices of at least one type must have been used within the first ten slices. – N. F. Taussig Feb 01 '21 at 10:30
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    @N.F.Taussig yes I thought about that but missed adding that part to the answer yesterday night. I will edit. Why some of these questions do not use words like at most two types remain or exactly two types remain is beyond me. – Math Lover Feb 01 '21 at 10:34
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    @N.F.Taussig thank you. that was a good callout. – Math Lover Feb 01 '21 at 12:01
  • Thanks all for the help! The exact wording was slices of only two of the three types remain, so I assume that to mean "exactly". For question 3, I am still a bit confused with the given answer. Would this just mean there are 10 ways to choose 2 out of 5 for each type, so 101010 ways for the numerator. Then, 5005 ways for the denominator? (choose 6 out of 15 randomly) – trevormil Feb 01 '21 at 13:34
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    @trevormil The answer for part 3 is indeed $$\frac{\binom{5}{2}\binom{5}{2}\binom{5}{2}}{\binom{15}{6}}$$ I do not think your interpretation that exactly two types of pizza remain in part 2 is correct for the reasons I stated above. – N. F. Taussig Feb 01 '21 at 13:37
  • Thanks! Yea I do see what you are saying. It is confusing wording, and I will try to clarify with the teacher. – trevormil Feb 01 '21 at 13:40