# Statistics probability questions

Yes, but with the 0.6 factor included, and using the correct choices of ##EX_1## and ##EX_2##.
Huh, I'm using that too along with 6.9 and 3.45, but I can't get that answer

Is the number you got 0.4539?

Ray Vickson
Homework Helper
Dearly Missed
Is the number you got 0.4539?

My p is (1-your p), so the answers agree when expressed in words.

Sorry for the confusion: I have been suffering from a cold that makes my head fuzzy, and so have been a bit mixed up.

StoneTemplePython
Gold Member
@Rifscape

To be honest, part (c) is probably one of the worst worded questions that I've seen. I hope you have a higher quality text to study from or is reviewing something like MIT's 6.041 (https://ocw.mit.edu/courses/electri...s-analysis-and-applied-probability-fall-2010/ ) or Harvard's intro to probability (see Joe Blitzstein on youtube). Otherwise there's a risk of learning the opposite of clear thinking from this course.

First criticism: see prior page where I pointed out that the problem is illogical if questions are during a given week for use in the following week, if we are talking about Stephane coming up with questions during the last week of the course. Based on results, it seems that the renewal idea is out the door, so my criticism that the question is busted is sustained. The fix would be to say that Stephane makes the decision to 'reward' students at the beginning of the second to last week of the course (so that the questions can be issued during the last week).

Second, and new criticism. The answer being sought violates the law of iterated expectations. Put differently the question asks one thing but wants an answer for a different problem. Let me simplify, and consider a simpler but probabilistically identical problem where instead of a student with a 60% chance of answering an exercise is considered, consider the case where top student with 100% chance of answering an exercise correctly is expected to answer 5 questions correctly.

The equation your professor wants is, have a prior distribution = ##\left[\begin{matrix}0.55\\0.45\end{matrix}\right]## and then use

##posterior \propto diag(likelihood) \left[\begin{matrix}0.55\\0.45\end{matrix}\right]
##

and your prof wants you to assume that 5 questions were issued -- and use that for your likelihood function.

##posterior \propto
\left[\begin{matrix}p(5, \mu= 3.45) & 0\\0 & p(5, \mu= 6.9)\end{matrix}\right]
\left[\begin{matrix}0.55\\0.45\end{matrix}\right]
##

##posterior \propto
\left[\begin{matrix}
0.12929992
& 0\\0 &
0.13135067
\end{matrix}\right] \left[\begin{matrix}0.55\\0.45\end{matrix}\right]
##

if your normalize the above (i.e. make sure the resulting vector sums to one), you get

##posterior =
\left[\begin{matrix}0.546102377853564\\0.453897622146436\end{matrix}\right]
##

- - - -
The issue is this is not what the question says. If the above is supposed to be the correct answer, then it should say,

Stéphane decides to "reward" the students by no longer limiting himself to 4 exercises, and instead assigning every exercise he writes. If the assignment is given out and it has 5 questions on it, what is the probably that Stéphane did not have a visitor that week?

or, more cumbersomely, it could say:
Stéphane decides to "reward" the students by no longer limiting himself to 4 exercises, and instead assigning every exercise he writes. The assignment is given out, and it has a certain number of questions on it, that is not a random variable. If a student with a 100% chance of correctly answering an exercise is expected to answer 5 questions correctly, what is the probably that Stéphane did not have a visitor that week?

but what it actually says (using my top student setup) is:
Stéphane decides to "reward" the students by no longer limiting himself to 4 exercises, and instead assigning every exercise he writes. If a student with a 100% chance of correctly answering an exercise is expected to answer 5 questions correctly, what is the probably that Stéphane did not have a visitor that week?

In this setup, the random variable is actually the number of questions on the test, not the probability of having a visitor -- the probability of a visitor is just a parameter we are estimating. Despite what the hint suggests, you simply cannot make the jump to there being a deterministic 5 exercises. The question must specify that the student already received the x number questions or in some other manner make it explicit that the number of questions is now fixed. As it currently reads, there is no reason to believe that the student has seen the questions (or that Stephane is done working on them). Instead what we in effect learn is that a bookie tells us, based on all publicly known information, a fair bet is that the top student will answer 5 questions correctly.

So the statement we actually get is far more general, and different, than being told that the top student has seen the homework assignment and that the top student expects to get 5 correct, and hence that there are only 5 questions being assigned (i.e. the number of questions being assigned is no longer a random variable).

The issue is that:

##posterior^T \left[\begin{matrix}3.45 \\ 6.9 \end{matrix}\right] =
\left[\begin{matrix}0.546102377853564\\0.453897622146436\end{matrix}\right]
^T \left[\begin{matrix}3.45 \\ 6.9 \end{matrix}\right] =
5.0159467964052062 \neq 5 ##

Hence the solution vector violates the law of iterated expectations, unless the student has seen the number of homework questions and determined it equals 5. The fundamental problem is that instead of telling you the number of questions assigned, the question writer (I believe, your prof) tried to get clever and give you an expected value statement that was not carefully worded. You may want to ask your prof about the law of iterated expectations, why the student's conditional expectation of questions being issued is not a random variable, and hence why said law of iterated expectations does not apply here. I would anticipate a fuzzy answer.

Ray Vickson
Homework Helper
Dearly Missed
@Rifscape

To be honest, part (c) is probably one of the worst worded questions that I've seen. I hope you have a higher quality text to study from or is reviewing something like MIT's 6.041 (https://ocw.mit.edu/courses/electri...s-analysis-and-applied-probability-fall-2010/ ) or Harvard's intro to probability (see Joe Blitzstein on youtube). Otherwise there's a risk of learning the opposite of clear thinking from this course.

Let me simplify, and consider a simpler but probabilistically identical problem where instead of a student with a 60% chance of answering an exercise is considered, consider the case where top student with 100% chance of answering an exercise correctly is expected to answer 5 questions correctly.

It makes perfectly good sense to ask for the posterior probability of a visiting scholar, given that a student answered 3 questions correctly and without making any assumption that the test has 5 questions. Admittedly the problem is more challenging than the "assume 5 questions" version, and may possibly be beyond the ability/knowledge of the OP, but I really do not know what the instructor intended, so having two possible versions cannot hurt. Below, let ##S## be the event "scholar week" and ##\bar{S}## the event "no-scholar week".

First of all, if 3 questions are answered correctly the test must contain ##N \geq 3## questions. For event ##\bar{S}## we have ##N \sim \text{Poisson}(\alpha)## with ##\alpha = 6.9,## while for event ##S## we have ##N \sim \text{Poisson}(\beta),## with ##\beta = 3.45##. Given an ##N = n \geq 3## the probability the student answers 3 questions correctly is the Binomial probability ##C(n,3) p^3 q^{n-3}, ## where ##p = 0.6, q = 0.4##. So, if ##Y## is the number of correctly-answered questions, then for event ##\bar{S}## we have ##P_{\bar{S}}(Y=3\: \&\: N=n) = C(n,3) p^3 q^{n-3} \alpha^n e^{-\alpha}/n!,## which simplifies to
$$P_{\bar{S}}(Y=3\: \&\: N=n) = \frac{p^3 \alpha^3}{3!} \frac{e^{-\alpha} (\alpha q)^{n-3}}{(n-3)!}.$$
Thus,
$$P(Y = 3|\bar{S}) = \sum_{n=3}^{\infty} P_{\bar{S}}(Y = 3\: \& \:N = n) = \frac{(\alpha p)^3}{3!} e^{-\alpha + \alpha q} = \frac{(\alpha p)^3 e^{-\alpha p}}{3!}.$$
Of course, this is Poisson probability with mean ##\alpha p = 0.6 \times 6.9 = 4.14.## Similarly, for event ##S##, ##Y## is Poisson with mean ##p \beta = 0.6 \times 3.45 = 2.07##. We have ##P(Y=3|\bar{S}) = 4.14^3 e^{-4.14}/3! \doteq 0.1883,## while ##P(Y=3|S) = 2.07^3 e^{-2.07}/3! \doteq 0.1865.##

The posterior probability of no visiting scholar, given that ##\{ Y = 3 \}##, is
$$P(\bar{S}|Y=3) = \frac{P(Y=3|\bar{S}) P(\bar{S})}{P(Y=3)} = \frac{(0.45)(0.1883)}{(0.45)(0.1883) + (0.55)(0.1865)}.$$

StoneTemplePython
First of all, if 3 questions are answered correctly the test must contain ##N \geq 3## questions.