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Question: A teacher gives $ 5 $ candies to $ 8 $ students. She gives each candy to a randomly chosen student, without regard to whether the student has received candy.

a) What is the probability that exactly $ 4 $ of the students get candy?

Here is my approach, I treat the candies to be identical and the students to be distinguishable: There are total of $ \displaystyle \binom{8 + 5 - 1}{5 - 1} = \binom{12}{4} $ ways to distribute $ 5 $ identical candies to $ 8 $ students. Next, there are $ \displaystyle \binom{8}{4} $ ways to choose $ 4 $ students to receive $ 5 $ candies so that each student among the $ 4 $ chosen must receive at least $ 1 $ candy, yielding a total of $ \displaystyle 4 .\binom{8}{4} $ possibilities. So the answer for part a) is $ \displaystyle \frac{4 .\binom{8}{4}}{\binom{12}{4}} $.


b) What is the probability that $ 5 $ different students get the candy?

I interpret this question as the same as: what is the probability that at least $ 5 $ students get the candy? Again, there are $ \displaystyle \binom{8 + 5 - 1}{5 - 1} = \binom{12}{4} $ ways to distribute $ 5 $ identical candies to $ 8 $ students. Next, I will find the number of possibilities that exactly $ 5, 6, 7, $ and $ 8 $ students get candies, respectively. There are $ \displaystyle \binom{8}{5} $ ways that exactly $ 5 $ students get candies. Note that there is no way more than $ 5 $ students get the candy if each student gets at least $ 1 $ candy. So the answer for part b) is $ \displaystyle \frac{\binom{8}{5}}{\binom{12}{4}} $.

Am I approaching the problem correctly? Any comment is appreciated.

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    $\begingroup$ Your answer to a) is unfortunately incorrect. You are correct in noting that there are $\binom{12}{4}$ ways to distribute the candies to the students, however those ways are not equally likely to occur. Perhaps it is easier to show this using b) instead. Give the candy out one at a time. For five students to get candy, give the first piece of candy to whoever. When you give the second piece of candy it must be to a different person than the first. When you give the third piece of candy it must be to a different person than the first to and so on, giving an answer of... $\endgroup$ – JMoravitz Apr 18 '17 at 22:03
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    $\begingroup$ (continued)... approximately $0.205$ (I leave the exact calculation to you to do) which is different than your answer of approximately $0.113$. It may help you to instead treat the candies as being distinct to see what is really going on. $\endgroup$ – JMoravitz Apr 18 '17 at 22:08
  • $\begingroup$ Ok so I follow your hint and I got $ \frac{8 \times 7 \times 6 \times 5 \times 4}{8^5} $ for part b). But I still can't see why my answer for part a) is incorrect. I can't use the same argument in part b) for part a) since the first, second, and third candy can be given to any of the $ 8 $ students. But the fourth and fifth candy depends on the first three candies. $\endgroup$ – newbie Apr 18 '17 at 22:11
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    $\begingroup$ Convince yourself that the stars-and-bars method of counting is incorrect by looking at a smaller example of two candies and two people and finding the probability that exactly two people get a piece of candy. By experimentation you should be able to agree the probability will be $\frac{1}{2}$, not $\frac{1}{3}$. For a correct calculation to part a), consider breaking into cases based on which candy in sequence happened to be the duplicate. $\frac{8\cdot 1\cdot 7\cdot 6\cdot 5+8\cdot 7\cdot 2\cdot 6\cdot 5+\dots}{8^5}$ $\endgroup$ – JMoravitz Apr 18 '17 at 22:14
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I would look at it this way, starting with b:

b) In order for five students to receive the five sweets, one student must receive one sweet each. The probability of the teacher selecting a different student each time equals:

$$\frac{8}{8} \cdot \frac{7}{8} \cdot \frac{6}{8} \cdot \frac{5}{8} \cdot \frac{4}{8} = \frac{6720}{32768} \approx 0.205$$

a) In order for four students to receive five sweets, one student must receive two sweets and three others must receive one. Let the teacher randomly select a first student. The probability of the teacher selecting this student again in the next turn equals $\frac{1}{8}$. If this happens, the teacher must select three different students afterwards, since one student already got two sweets. If this does not happen (i.e. two different students were selected), there are two options in the third turn: one of the two students who already got a sweet are selected, or a new student is selected. The former case occurs with a probability of $\frac{2}{8}$, in which case two different students must be selected afterwards. If the teacher however selects a third unique student, the probability of the teacher selecting one of them on the fourth term equals $\frac{3}{8}$ (in which case a different student must be selected in the last term). If a new student is however selected, one of the four selected students must be selected again in order to receive two sweets, which happens with a probability of $\frac{4}{8}$. All in all, the probability is thus:

$$\frac{8 \cdot 1 \cdot 7 \cdot 6 \cdot 5}{8^5} + \frac{8 \cdot 7 \cdot 2 \cdot 6 \cdot 5}{8^5} + \frac{8 \cdot 7 \cdot 6 \cdot 3 \cdot 5}{8^5} + \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{8^5} \approx 0.512$$

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