Statistics probability questions

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Homework Statement



Each week, Stéphane needs to prepare 4 exercises for the following week's homework assignment. The number of problems he creates in a week follows a Poisson distribution with mean 6.9.

a. What is the probability that Stéphane manages to create enough exercises for the following week's homework? Round your answer to 4 decimal places.

b. Unfortunately, each week there is a 55% chance that a visiting scholar from Switzerland arrives and burdens Stéphane with research questions all week. During these weeks he only writes an average of 3.45 exercises. If Stéphane fails to write 4 exercises one week, what is the probably that he received a visiting scholar that week? Round your answer to 4 decimal places.

c. The last week of the semester, 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 60% chance of correctly answering an exercise is expected to answer 3 questions correctly, what is the probably that Stéphane did not have a visitor that week? Round your answer to 4 decimal places.
Hint: First find the number of exercises in the last week of the semester from the chance and expected value of the correct answers.

Homework Equations


P(A and B) = P(A) * P(B)
P(A | B) = P(A and B)/p(B)
Poisson distribution equation:
P(x; μ) = (e-μ) (μx) / x!

The Attempt at a Solution



I was able to finish the first question and get the right answer but I'm having trouble on parts b and c.
For the first question:
P(x = 4) = 1 - P(x <= 3) = 1 - (P(0) + P(1) + P(2) + P(3))

Then I used the poisson distribution equation and was able to get the answer.

I think for b you need to use the conditional equation, but I'm not sure what the P(failing) would be.

I have no idea how to do c

Any help is appreciated, thanks for reading
 

Answers and Replies

  • #2
StoneTemplePython
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The formatting here is kind of hard to read... I think your Poisson distribution equation is wrong though. If you are using parameter ##\mu ## instead of the perhaps more typical ##\lambda##, the poisson pmf is given by:

##P(x, \mu) = \frac{\mu^x e^{-\mu}}{x!}##

I'd focus first on part b. Draw a tree with 55% chance of having a situation with ##\mu := 3.45## and a 45% chance of ## \mu = 6.9##. Trees are visually powerful, and quite useful in computation settings -- so give it a shot and draw this tree. (Historical note: drawing a tree is also how Pascal solved the 'original' probability problem -- the problem of points.) For each of the two leaves of this simple tree, what is that probability of having not done at least 4 exercises? I.e. what is ##\big(P(x=0, \mu) + P(x=1, \mu) + P(x=2, \mu) + P(x=3, \mu)\big)## for both leaves of this tree? From here you just need to get a normalizing constant so that your 'posterior' sums to one. Are you familiar with this?

Part C flips the conditioning around -- ala Bayesian inference, but I'd focus on solving b first.
- - - -

btw for part A), your equation says "P(x = 4) " but it really should read ##P(x \geq 4)##
 
  • #3
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Alright I did that, and was able to get the P(failing) since P(failing) = P(failing|scholar) + P(failing|no scholar)

Alright I was able to get .88 for the second question, though I don't know how to do the last part.
 
  • #4
Ray Vickson
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Alright I did that, and was able to get the P(failing) since P(failing) = P(failing|scholar) + P(failing|no scholar)

Alright I was able to get .88 for the second question, though I don't know how to do the last part.
No, that is not correct; the correct result is
P(fail) = P(fail|scholar)*P(scholar) + P(fail|no scholar)*P(no scholar).
 
  • #5
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Wait yeah you're right, I did that, just forgot to write it down. You can only get 0.88 if you do P(fail) = P(fail|scholar)*P(scholar) + P(fail|no scholar)*P(no scholar).

Do you know how to start part c?

Thanks for the help
 
  • #6
StoneTemplePython
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Shouldn't (c) be something like ##0.6(p*E[X_1] + (1-p)*E[X_2]) = 3##

via law of total expectation and linearity of expectations? All you are doing is solving for p, there. Again, drawing a picture first -- i.e. a tree-- should get you there.

I really can't emphasize this enough -- if at all possible, try drawing picture to help solve your problems in probability. You should pretty much always be able to do this when you are working with a countable number of states...
 
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  • #7
Ray Vickson
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Wait yeah you're right, I did that, just forgot to write it down. You can only get 0.88 if you do P(fail) = P(fail|scholar)*P(scholar) + P(fail|no scholar)*P(no scholar).

Do you know how to start part c?

Thanks for the help

I don't get 0.88 for part (b).
 
  • #8
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Yeah I drew a tree for part b and it made sense and I was able to get it, but I have been unable for this question. I attempted to draw one based on the formula you gave, but I don' think its fully correct.

When I plugged in 6.9 for E[X1] and 3.45 for E[X2], I solved for p and got p= 0.4492, which is not correct. Though I think I am on the right track.
 
  • #9
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I don't get 0.88 for part (b).

I got that value after dividing the P(scholar arriving and failing) by the product of (fail) = P(fail|scholar)*P(scholar) + P(fail|no scholar)*P(no scholar).
 
  • #10
StoneTemplePython
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I don't get 0.88 for part (b).

I got that value after dividing the P(scholar arriving and failing) by the product of (fail) = P(fail|scholar)*P(scholar) + P(fail|no scholar)*P(no scholar).

Setting aside the nit that the answer wants 4 decimals, 0.88 checks out and is correct.
 
  • #11
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Alright cool thats good, just need to solve part c then. The p value I get from that equation doesn't seem to work.

Would X1=6.9 and X2=3.45? Or are the expected values different from the means of a poisson distribution
 
  • #12
StoneTemplePython
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Alright cool thats good, just need to solve part c then. The p value I get from that equation doesn't seem to work.

Would X1=6.9 and X2=3.45? Or are the expected values different from the means of a poisson distribution
To be clear ##X_1## and ##X_2## are random variables. But HIGH LEVEL (see bold at end) yes, you might interpret it as ##E[X_1] = \mu_1## and ##E[X_2] = \mu_2##. Unfortunately a lot of probability problems turn into linguistic ones where meanings easily get trampled on...

The issue I have from C is the question has linguistic issues.

Your problem begins by saying

"Each week, Stéphane needs to prepare 4 exercises for the following week's homework assignment."

Then part c says:

The last week of the semester, Stéphane decides to "reward" the students by no longer limiting himself to 4 exercises

I have two interpretations here. One the question is busted -- the idea of Stéphane preparing questions for the following week during the last week simply doesn't make any sense, as those questions would be issued after the semester end. The alternative interpretation is -- loosely speaking-- that we have a renewal here -- so basically you have a Poisson with a cap at 4 from prior week, and then another unbounded Poisson that happens during the last week for issuing more exercises. So now you'd draw a tree with a root, and two levels in it. That's my current thinking at least -- the wording has room for improvement.
 
  • #13
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Yeah, my professor always has these wording issues. I think its probably the second option, where there is a renewal and that there is a new cap.
 
  • #14
Ray Vickson
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I got that value after dividing the P(scholar arriving and failing) by the product of (fail) = P(fail|scholar)*P(scholar) + P(fail|no scholar)*P(no scholar).

You are correct; I was calculating the wrong thing.
 
  • #15
Ray Vickson
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Alright cool thats good, just need to solve part c then. The p value I get from that equation doesn't seem to work.

Would X1=6.9 and X2=3.45? Or are the expected values different from the means of a poisson distribution

The number of questions answered correctly by the student is neither ##EX_1= 6.9## nor ##EX_2 = 3.45##. These quantities are the numbers of questions on the quiz paper, (with or without a visiting scholar) but on average the student answers some of them incorrectly.
 
  • #16
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The number of questions answered correctly by the student is neither ##EX_1= 6.9## nor ##EX_2 = 3.45##. These quantities are the numbers of questions on the quiz paper, (with or without a visiting scholar) but on average the student answers some of them incorrectly.

How would you get the E[X1] and E[X2] then? I'm still not really sure how I would tackle this question, I drew a tree, but there seems to be many parts to this question.
 
  • #17
StoneTemplePython
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The number of questions answered correctly by the student is neither ##EX_1= 6.9## nor ##EX_2 = 3.45##. These quantities are the numbers of questions on the quiz paper, (with or without a visiting scholar) but on average the student answers some of them incorrectly.

@ Ray Vickson
OP was responding to my equation from earlier where I said: ##0.6(p*E[X_1] + (1-p)*E[X_2]) = 3##


How would you get the E[X1] and E[X2] then? I'm still not really sure how I would tackle this question, I drew a tree, but there seems to be many parts to this question.

@ Rifscape
I would still recommend using my above setup. I don't understand why Ray Vickson would make such a comment.

If you want to formalize this and abstract even further --- note: the following is unnecessary-- you can use the law of iterated expectations. Consider some random variable ##Y## which denotes the number of questions answered correctly by the student. Consider ##N##, a natural number denominated r.v. that gives the number of questions actually offered.

Consider Q, a bernouli r.v. that has probability 0.6 of a correct answer by said student for a given question and is independent of ##N##.

The question states ##E[Y] = 3##. Using law of iterated expectations we can rewrite this as

##E[Y] = 3##
## E[Y] = E[E[Y|N]] = 3 ##
## E[Y] = E[N E[Q]] = 3 ##
## E[Y] = E[N] E[Q] = 3##
## E[Y] = E[N] (0.6) = 3##
## E[Y] = 0.6 E[N] = 3##


When you consider that ##E[N] = (p*E[X_1] + (1-p)*E[X_2])##, you then get the equation I originally supplied. I am not supposed to supply the whole answer, so I leave finding ##E[X_1]## and ##E[X_2]## up to you, though maybe I can help if you have some follow-up questions where I won't have to give away the whole thing.
 
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  • #18
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I'm still kind of stuck on how to find E[X1] and E[X2], would I use the poisson distribution equation and find the probability that x >=3 using 6.9 for X1 and 3.45 for X2?
 
  • #19
Ray Vickson
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I'm still kind of stuck on how to find E[X1] and E[X2], would I use the poisson distribution equation and find the probability that x >=3 using 6.9 for X1 and 3.45 for X2?

In the original problem statement it said "The number of problems he creates in a week follows a Poisson distribution with mean 6.9" before part (a) and said "During these weeks he only writes an average of 3.45 exercises" in part (b). What do you think those statements signify?
 
  • #20
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In the original problem statement it said "The number of problems he creates in a week follows a Poisson distribution with mean 6.9" before part (a) and said "During these weeks he only writes an average of 3.45 exercises" in part (b). What do you think those statements signify?
Wouldn't those be the expected values, for x1 and x2, since it's a poisson distribution? But I already tried those values and they didn't work.
 
  • #21
Ray Vickson
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Wouldn't those be the expected values, for x1 and x2, since it's a poisson distribution? But I already tried those values and they didn't work.

They should have worked. When I solved for ##p## I get a value different from your 0.4492.
 
  • #22
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@ Ray Vickson
OP was responding to my equation from earlier where I said: ##0.6(p*E[X_1] + (1-p)*E[X_2]) = 3##
I got x1 as 5 and x2 as 3, but when I plug them in I get?



@ Rifscape
I would still recommend using my above setup. I don't understand why Ray Vickson would make such a comment.

If you want to formalize this and abstract even further --- note: the following is unnecessary-- you can use the law of iterated expectations. Consider some random variable ##Y## which denotes the number of questions answered correctly by the student. Consider ##N##, a natural number denominated r.v. that gives the number of questions actually offered.

Consider Q, a bernouli r.v. that has probability 0.6 of a correct answer by said student for a given question and is independent of ##N##.

The question states ##E[Y] = 3##. Using law of iterated expectations we can rewrite this as

##E[Y] = 3##
## E[Y] = E[E[Y|N]] = 3 ##
## E[Y] = E[N E[Q]] = 3 ##
## E[Y] = E[N] E[Q] = 3##
## E[Y] = E[N] (0.6) = 3##
## E[Y] = 0.6 E[N] = 3##


When you consider that ##E[N] = (p*E[X_1] + (1-p)*E[X_2])##, you then get the equation I originally supplied. I am not supposed to supply the whole answer, so I leave finding ##E[X_1]## and ##E[X_2]## up to you, though maybe I can help if you have some follow-up questions where I won't have to give away the whole thing.
 
  • #23
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They should have worked. When I solved for ##p## I get a value different from your 0.4492.
Hmm really what value ? I keep getting 0.4492, is it the same equation that python gave?
 
  • #24
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They should have worked. When I solved for ##p## I get a value different from your 0.4492.
Isn't it (3/0.6 - 3.45) /3.45, which equal. 4492
 
  • #25
Ray Vickson
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Hmm really what value ? I keep getting 0.4492, is it the same equation that python gave?

Yes, but with the 0.6 factor included, and using the correct choices of ##EX_1## and ##EX_2##.
 

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