The university you are attending is holding a huge fundraiser so every student, including yourself, is going. At the even you are seated with three other students per table. The problem is many of the students on campus you don't know so your hoping the people you sit with you will know. The dorm you live in has .02 of the total student population. You know .4 of the students living in your dorm, but you know only .05 of the rest of the students on campus. The number of students is large, so assume that the events of knowing different students at the fundraiser are independent.

The following questions are posed.

For $i = 0, 1, 2, 3$, what is the probability of you sitting with $i$ people that you know?

Given that $i$ students (where $i = 0, 1, 2 or 3$) from your dorm are sitting at your table, what is the probability that you know all the people at your table?

The host tells you that you are seated at a table with 3 other people you know. What is the probability that all these three students at your table are from your dorm?

So the probability of knowing someone at your table is (.02)(.40)+(.98)(.05)=.057. So to answer question one we need a series so I did the following:

(.057)$\sum_{i=0}^{3}$$3 \choose i$=$(.057)(1+3+3+1)$=$.456$

For the second question I was not sure but I did something similar to the first question.

$\sum_{i=0}^{3}$$3 \choose i$$(.008)^i$=(8)(1.00808) which clearly doesn't make sense so I am not sure what I am doing wrong for that.

For the third question since we calcualted the probability we know someone at the table I divided that by the chance you know someone from your dorm to get, $\frac{(.02)(.4)}{.057}$=.140351. Since we are talking about all three I did $.140351^3$=.002765

I think I did the first and third questions correctly but the second one I def did not.

  • $\begingroup$ I don't think you did the first question correctly. It reads, to me at least, like four similar problems: What is the P of knowing 0 of the other 3 at the table? ... of knowing 1? ... of knowing 2? [and]... of knowing all 3? $\endgroup$ – The Chaz 2.0 Feb 13 at 15:13

We make the assumption that the total number of people are "large" and so we may model our approach on selection with replacement rather than without.

Now... let us look at the probability that we know a very specific person.

You correctly found that this will be $.02\cdot .4 + .98\cdot .05=.057$

However, you have misinterpreted or misused this value as it appears in later calculations. You incorrectly used this as the probability that you know all three students at a table rather than how you should have used it which is the probability that you know a specific single student without regards to whether or not you know any others.

Now... the probability of knowing $i$ of the three people sitting with you at your table, we can model with a binomial distribution. To know a specific person occurs with probability $.057$ and to not know a specific person, this occurs with probability $.943$.

To know none of the people at your table, i.e. $i=0$, this occurs when you don't know the first person, you also don't know the second person, and we also don't know the third person. Thanks to our assumption on the number of people being "large" we can find the probability of all of these things happening simultaneously as being the product of these values, $(.943)^3$.

To know exactly one of the people at your table, i.e. $i=1$, this occurs when you don't know the first person or the second person but you do know the third, or you don't know the first and third person but do know the second, or you do know the first but not the second or third. Each of these values we find will be the same, so we can add the three of them together which can be written more simply as multiplying by three and will be $3\cdot (.943)^2(.057)^1$

More generally, we can use the binomial distribution to write this much more quickly and generally, letting $X$ be the random variable counting the number of people at the table we know, that $X=i$ as:


On to the next question, given that $i$ of the students at your table are dorm mates of yours, we are curious to find the probability that we know everyone at the table.

Well... with $i$ people that we are dorm mates with, we want to find the probability that we know each individually, and for the remaining $3-i$ people we want to find the probability that we know of them individually, and multiply the probabilities together.

$$(.4)^i\cdot (.05)^{3-i}$$

The third problem you should approach with Bayes' theorem and will wait to see if you understand what I've already written before continuing too far, but will involve you needing to calculate various other probabilities such as the probability that all three people are members of your dorm and you know all three, finding the probability that you know all three, and so on...

For the third: letting $A$ be the event that we know all three people and $B$ the event that all three people are dorm-mates of ours, we were tasked with calculating $Pr(B\mid A)$

$$Pr(B\mid A)=\dfrac{Pr(A\cap B)}{Pr(A)} = \dfrac{Pr(B)Pr(A\mid B)}{Pr(A)}=\dfrac{(.02)^3(.4)^3}{(.057)^3}$$

  • $\begingroup$ For the first question I got all my values to add to one so I am assuming I am correct. For the second question I followed the format you layed out so would I just be doing $i=3$, $(.05)^3$ since we are looking for knowing 3 students? Finally for the last question, I did the probability of knowing each for the second question so would it be $\frac{(.4)^3 }{(.05)^3+(.4)(.05)^2+(.4)^2(.05)+(.4)^3}$=.861953. Is this approach correct or no? $\endgroup$ – chero21 Feb 13 at 17:47
  • $\begingroup$ @chero21 no, there are several errors. I already gave you the answer to the second problem. In this context $i$ is not in reference to how many students we know but is instead in reference to how many people at the table are dorm-mates of ours. Just like in the first, we are interested in each of the possible final results for each of the possible values of $i$. $\endgroup$ – JMoravitz Feb 13 at 17:52
  • $\begingroup$ @chero21 for the final problem, your denominator should have been the answer to the first problem in the case of $i=3$., it should have been the probability of knowing all three students. Then the numerator should have been the probability of all three students being dorm-mates of yours and them all being people you know. Your denominator in your attempt was the probability that all of them are known given that they were all dormmates, something very different. $\endgroup$ – JMoravitz Feb 13 at 17:56
  • $\begingroup$ so for the second problem $i=3$ was one of the many solutions to this problem since that we know all 3 people. Once I did $i=0,1,2$ I got values for all three but they did not add up to 1, but I believe it's the same circumstances as the first question. As for the third question, if I am understanding you correctly it would be $\frac{(.000185)(.064)}{.00185}$ which give us .064. $\endgroup$ – chero21 Feb 13 at 18:20
  • $\begingroup$ @chero21 we don't need for the answers in the second problem to add up to $1$. We don't expect them to either. We expect events which partition the sample space to have probabilities that add up to $1$. We expect events which partition another event to have conditional probabilities conditioned on that other event that add up to $1$. Here for the second problem, we have neither of these. Rather, what we are conditioning on are a partition of the sample space, something very different. $\endgroup$ – JMoravitz Feb 13 at 18:28

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