The simple answer is: the expectation of the norm is not the same as the norm of the expectation .The norm is not a linear function, so the E[f[X] ] != f[ E[X ]]. So this should work instead:

Expectation[Norm[{x,y}],{x,y}\[Distributed]UniformDistribution[2]]

Where can I download the updated study materials? I found it at https://amoeba.wolfram.com/index.php/s/LgfWrztXo9xELo4. Are these the notebooks that have been updated with the corrections mentioned in this community post? I noticed that the most recent modification dates were set to 2 months ago.

Hi Peter,

Indeed, as Dave said, please refer to the ressource found on Wolfram U for the latest version of any notebook.

Hello Dave,

Thank you for this exhausive review, this is currently being addressed and will be changed soon on the framework. Let me address these concerns one by one for you:

Lesson 2: correct, unique objects are used.

Exercice 7: correct, the function in the question will be modified.

Exercice 10: correct, the + are now added.

Exercise 15: correct, the coefficients will be added.

Exercise 18: 4: correct, the word will be changed. 5: incorrect, this is not a root, the only root taken is because of the transfer from variance to standard deviation.

Exercise 19: this will not be changed, since this is a question of interpretation of the dataset, a skill we want to develop in this course.

Exercise 21: slight changes were made to make it clearer. However, consider the only requirement here is to try to interpret clusters. Moreover, answers only always portray one possible way to answer. Giving steps would go against the many possible ways to answer the question.

Exercise 23: This was already corrected from another community post. Moreover, the -1 is not necessary, only that it is a negative number. There is no issue if your method is slightly different. Question 4: both are true. The questionnaire goes from 1 to 10, but the distribution is from 3 to 10. in other words, the probability of 1 to 2 is 0, as no one gave that answer. Question 5: this is found in the term "average" of distributions. A combination of distributions does not imply equal weights, but an "average" does.

Lesson 24: this seems to be a framework problem, this will be updated soon.

Exercise 24: No, there is no mathematical difference. You are adding 0.5, is 0.5 included or excluded? Not only does this make no difference in calculation, it also is impossible to argue one or the other semantically, as the approximation is independent of a finite point on an interval. So in the continuous case, it is preferrable to avoid equality symbols since it complicates the formulation without changing anything, even if we're aiming for the most absolute mathematical correctness.

Quiz 6: this was corrected.

Practice Exam: correct, this will be corrected soon.

Thank you again for this huge help in the review of this course. Good luck on your probabilistic endeavors!

Hello Mitchell,

For all errors or bugs, it takes us about 1 or 2 days for it to be corrected and on the course framework. So to access the updated version, always refer to the notebooks found on the course framework. The practice exam and multiple exercise notebooks were corrected since the launch of this course, so please redownload those on the course framework if you were using the old original notebooks.

Hello Joseph,

Indeed, this seems to be a mistake, it will be corrected. Thank you for noticing.

Hello Joseph,

As said in the question, there are many approaches to the problem. You may set your standard deviations in whatever way you want. Maybe the only restriction is that it doesn't surpass the standard deviation of exercise 4:

N@StandardDeviation@DiscreteUniformDistribution[{3, 10}]

which is about 2.29. The key here is to experimentally explore the difference in convergence between the weak and strong law of large numbers.

Hello Joseph,

Indeed, this doesn't make sense. This mistake was caught and corrected, please refer back to the framework exercise notebooks.

That makes a lot more sense now, thanks!

Hello Joseph,

Thank you for noticing, this will be corrected shortly.

Hi John,

This is now fixed, thank you for noticing.

Also for Exercise 5 in the same lesson:

Nest[TransformedDistribution[
   x + y, {x \[Distributed] #, y \[Distributed] #}] &, 
 NormalDistribution[2, Sqrt@31/2], 2]

Hello,

We will correct this shortly, I'll post again to confirm it has been corrected.

In question 35 of the Practice exam, we consider the permutations of a group of letters to determine which are words. It seems that Mathematica finds an extra word if the letters are capitalized….

In[1]:= Tally[DictionaryWordQ/@StringJoin/@Permutations[{"r","e","s","e","t"}]]

Out[1]= {{True,6},{False,54}}

In[2]:= Tally[DictionaryWordQ/@StringJoin/@Permutations[{"R","E","S","E","T"}]]

Out[2]= {{True,7},{False,53}}

The word “STERE” is missing from the uncapitalized words...

On Exercise 1 of Exercises-23.nb, I believe the solution shows the upper bound for the probability that x<=13 or x>=37. Then 1 minus that probability gives a lower bound for P(13<=x<=37). The same thing happens in Problem 2 of Quiz 6.

Also, for Problem 5 in Quiz 6, one answer shows the probability of x>260 using a Normal to approximate Discrete Data, but no answer shows for x>260+.5.

On Exercises 1 and 2 from Exercises24.nb, the solutions use the variance instead of the standard deviation as the second parameter for NormalDistribution[].

On exercise 2 of Excercises25, the solution uses different attributes.

Hello Juan,

1. This is new and will be corrected, indeed it should be a lower bound, 1-the probability.
2. Why would this be +0.5? "At least" implies inclusion, so -0.5 is more appropriate.
3. Right on! We will correct this! Personally this is one of my most reoccurring mistakes, I find it very counterintuitive to use standard deviation as a parameter instead of variance. Thank you for catching that.
4. Definitely a small but significant mistake. We will correct this.

Thank you a lot of all those corrections!

Hello Joseph,

There seems be two problems mixing here, so this will be corrected to:

The distribution of values of the retirement package offered by a company to new employees is modeled by the probability density function 1/5 e^(-(1/5)(x-5)) for x>5. Calculate the variance of the retirement package value, given that the value is at least 10.

The code would be:

Expectation[x^2 \[Conditioned] x > 10, 
  x \[Distributed] retirementDist] - 
 Expectation[x \[Conditioned] x > 10, 
   x \[Distributed] retirementDist]^2

This results in the same variance however, since this is condition independent.

Thank you for noticing.

Hello Mitchell,

1. The first is true, the second should be

   P(A' \[Union] B') = P(A') + P(B') - P(A' \[Intersection] B')
   

or

    1- P(A  \[Intersection] B) = P(A') + P(B') - P(A' \[Intersection] B')

Could you tell me where you found this so that I may correct it?

1. Yes, you can assign any functionality to a weighted dataset with WeightedData, for example:

   data = RandomReal[{-5, 5}, 10^4];
   weightedData = WeightedData[data, PDF[NormalDistribution[], #] &]
   EstimatedDistribution[weightedData, 
    NormalDistribution[\[Mu], \[Sigma]]]
   

Some functionalities, like FindDistribution, may not work with WeightedData, but most functionalities do.

1. As mentionned when introducting FindDistribution, that function is only for univariate data. You may want to FindDistribution for each dimension (independence assumption) or use for example a general multinormal distribution with EstimatedDistribution (normal assumption, but dependence is allowed).

I'm glad you enjoyed the course, I wish you the best!

Hello Mitch,

In two words: De Morgan's!

P(A' \[Union] B') = P((A'' \[Complement] B'')') (De Morgan's Law of Set Theory)
P(A' \[Union] B') = P((A \[Complement] B)') (Double Negation of Set Theory)
P(A' \[Union] B') = 1 - P(A \[Complement] B) (Complement Law of Probability Theory)

Those laws are always useful, once in a while. I hope that helps.

Hello,

If you are referring to the mock exam, I can't see anything missing. If you are referring to the actual Final Exam, the numbering is random, so I will need more information to be able to pinpoint the question.

The course materials have been updated with the corrections :)

Hello,

The lastest notebooks are on the course framework, or soon to be on the framework. If you find an error, just ctrl-find it on this page or suggest a change if you can't find it. But most mistakes were corrected in the framework by now. Hopefully that helps.

Hello Juan,

As repeated throughout the course, it all comes down to your capacity to recognize where to apply each distribution. If you can recognize the situation is appropriate for a specific distribution, use EstimatedDistribution on that abstract distribution together with the data. Then extract your measures from that distribution. Now, if you only have data and no information otherwise, use FindDistribution and actually use the found distribution to take your measures.

Hello Juan,

I'm not sure how you are getting that result. The output for:

EstimatedDistribution[{1,3,6,0,3,4,5,7,4,2,11,1},PoissonDistribution[\[Lambda]]]

is

PoissonDistribution[3.91667]

Hello Michael,

Indeed! This may see weird at first but this comes down to the approach. If you know what shape you're facing for the PDF, then it becomes a problem of classical probability theory to find the parameters corresponding to that PDF. However, FindDistribution is much more complex and based on heuristics, which make it much more uncertain. It tries to imagine the data that is not there, using heuristics. For example, with minimal data, it will prefer the Uniform and Normal distribution as they are common, even when the shape is different. This is why the course tends to emphasize EstimatedDistribution, whenever you have just a little bit of info more than raw data.

Hello Ahmed,

Indeed, you also need to be careful about that. When a parameter is not used in the function signature, it's usually because that parameter is assumed to be some standard value, like 0 or 1. If you cannot make such an assumption in your situation, it is in your interest to use the most general form of the distribution.

Hello Juan,

The normal distribution is the distribution of approximations, it is used to approximate two major distributions: the binomial and the Poisson distribution. In both approximation, just use the mean and variance of the exact distribution you're trying to approximate. If it's the binomial, take the binomial mean and variance for the normal. If it's Poisson, take the Poisson mean and variance.

For exercises08 exercise 2 , given as "What is its expectation of the function Binomial[10,i]/2^10 for 0<=x<=10?" What is the "its" referred to in the problem statement ? The solution seems to be calculating the expectation of x for the distribution Binomial[10,i]/2^10 .

I also get a different output from Lesson 14 Exercise 4 with the same input as the exercise gives:

EstimatedDistribution[{3, 3, 10, 6, 6, 4, Sequence[
  5, 9, 3, 4, 7, 4, 7, 10, 8, 5, 6, 7, 11, 10, 5, 9, 7, 8, 6, 5, 6, 7,
    6, 8, 12, 9, 6, 3, 9, 5, 7, 5, 2, 9, 3, 5, 9, 9, 3, 5, 3, 8, 5, 6,
    5, 4, 7, 10, 6, 7, 8, 8, 11, 9, 8, 8, 9, 3, 11, 8, 7, 10, 5, 4, 5,
    10, 4, 8, 7, 7, 4, 3, 5, 10, 5, 4, 11, 5, 6, 10, 5, 7, 10, 11, 7, 
   5, 4, 7, 9, 5, 4, 5, 7, 5, 10, 11, 10, 5, 5, 7, 4, 7, 5, 4, 3, 4, 
   7, 10, 4, 8, 2, 7, 4, 4, 8, 4, 8, 8, 3, 9, 7, 7, 7, 7, 10, 5, 9, 8,
    11, 6, 8, 7, 7, 8, 3, 6, 7, 6, 7, 8, 8, 7, 2, 3, 4, 9, 7, 7, 6, 4,
    10, 6, 4, 8, 10, 7, 3, 10, 6, 6, 6, 5, 9, 7, 11, 6, 7, 1, 4, 8, 8,
    5, 5, 2, 8, 6, 7, 7, 5, 5, 6, 5, 6, 2, 12, 7, 6, 5, 7, 5, 9, 6, 4,
    8, 3, 8, 3, 7, 6, 3, 10, 6, 3, 6, 7, 8, 7, 3, 7, 4, 5, 4, 10, 8, 
   7, 10, 10, 7, 5, 9, 5, 4, 6, 4, 6, 11, 7, 9, 9, 6, 7, 4, 6, 7, 5, 
   5, 5, 5, 6, 4, 8, 4, 8, 7, 6, 4, 4, 5, 7, 8, 4, 2, 1, 5, 9, 2, 6, 
   11, 5, 4, 5, 12, 7, 7, 7, 0, 3, 7, 4, 6, 11, 5, 3, 5, 8, 4, 5, 2, 
   3, 8, 8, 6, 6, 1, 9, 4, 3, 8, 5, 4, 4, 5, 4, 5, 6, 6, 5, 7, 6, 1, 
   7, 3, 9, 8, 4, 8, 2, 9, 7, 13, 5, 5, 2, 8, 12, 8, 5, 5, 2, 3, 4, 9,
    11, 5, 6, 10, 5, 5, 5, 6, 5, 4, 3, 8, 8, 12, 7, 7, 8, 11, 2, 9, 
   10, 5, 4, 2, 8, 9, 6, 8, 7, 6, 1, 8, 7, 9, 10, 5, 10]}, 
 PoissonDistribution@\[Mu]]
Variance@%

PoissonDistribution[6.31507]
6.31507

The values used in Lesson 14 Exercise 2 do not match the values given in the question. For that question I get the following input and output:

Probability[x >= 25, 
  x \[Distributed] PoissonDistribution[0.06 500]] // N
Probability[x >= 25, x \[Distributed] BinomialDistribution[500, 0.06]]

0.842758
0.850619

Is this correct?

Hello Juan,

All correct! Thank you for noticing.

1. Indeed, your answer makes more sense given the formulation. It will be corrected.
2. Indeed, the formulation is too ambiguous, it will be corrected to make it more clear what is needed.
3. Yes, for exercise 2 and 4, there are some equalities missing to the final solution.

Thank you!

Hello Joseph,

This is indeed also correct, and is an equivalent formulation of:

Probability[x < 19 \[Conditioned] x >= 17, x \[Distributed] dist]

As mentioned previously in this post.

Hello Parker,

Indeed, it seems the question and solution numbers don't correspond. This will be corrected.

Thank you for noticing.

Marc: I don't quite understand your statement "as soon as I have three balls the game is over". I did assume that once "you" draws a red ball the game is over. I reproduced the terms in your sum but if you carry the tree to the end you get some cases where there are 3 balls left and "you" have not yet drawn a red ball. I work this out in the attached notebook

Thanks for your response. I could not help thinking that there must be an easier way to solve this problem than by drawing a complex tree!

Hi Marc,

I am reviewing "Mock Exam" notebook and have found that the Question 31 solution is not correct. Given U, the transformed distribution with U-0.3 must be stated as:

Plot3D[{PDF[DirichletDistribution[{5, 3, 2}], {x, y}], 
  PDF[TransformedDistribution[{a - 0.3, 
     b - 0.3}, {a, b} \[Distributed] 
     DirichletDistribution[{5, 3, 2}]], {x, y}]}, {x, 0, 1}, {y, 0, 
  1}, Filling -> Bottom, PlotRange -> All, 
 PlotLegends -> {"U", "U-0.3"}]

Hi Marc, It seems there are problems in both Quiz 6 #1 and #6, both of which are calculating the probability by normal approximation to binomial distribution. Please check these two questions in the "framework".

Attachments:

Normal-Approximation.nb

Thanks for this explanation, Marc. I have tried again using N@Probability. The 1st question still indicates that my response is wrong. As for the 6th question, it seems that there is no correct answer in the multiple choice. I suspect that there is either a technical glitch or wrong syntax in the "Framework". Please do the double check. Best wishes,

Can you clarify why you chose x>=440-0.5 for the argument for the Probability statement in your analysis of Exercise 6. I certainly see why 440 is chosen but I don't get the -0.5.

Response

The question mentions to use if appropriate the normal approximation. Read or listen again to Lesson 24 to know how to apply that approximation. Here 440 is included, so we include it in the probability.

I will look this over.

Hi Marc, Can you quickly check whether there is any technical error in your courseware ("framework")?

I am sorry. I figured out my mistake. The correct syntax is:

NProbability[x > 40, 
 x \[Distributed] BinomialDistribution[150, 0.2633]]

Attachments:

Quiz-3-3 Question.nb

Thanks for making the point about spring break, @Dave Middleton. We will extend both deadlines by a week, so quizzes should be done by March 24, and the exam by March 31. Yes, you can earn both completion and Level 1 independently in the interactive course, but course completion will require that you watch the video lessons within the framework. We run custom data pulls in order to verify completion with the Study Group.

Hi;

When you obtain the probability, it is easy to understand its exact meaning. However, the Expectation and RandomVariate are not so easy to understand. When would you use these two and what exactly are they telling me?

I have been puzzled about conditional probability, because is the situation usually as clear as that? The probability of an event occurring, given that another event has already occurred. It seems to me that in real world situations there can be all kinds of complex correlations between the events and the external world, not captured by the basic conditional probability formula.

Interesting, thanks.

Hello Hakan,

Indeed, the version on the framework seems to be the wrong one! Thank you for telling us, this will be corrected shortly.

Hello Joseph,

First, since we aim to fit every phrase in a single line, we tend to shorten definitions like this one in interest of simplicity, while maintaining correctness.

Second, I think I actually disagree with your definition.

1. Set theory is more basic than probability theory, that much should be clear. Thus, set operation definitions should not be dependent on the definitions of probability theory of event and sample space. So the vocabulary is, dare I say, anachronic.

2. In set theory, your confusion is actually correct. Due to the trivial existence of the theoretical universal set, the complement of a given set can be used despite not referring to any exterior set. It's just another variable, that may or may not contain every possible element or not. The generality of the statement doesn't have to be lost. So in that sense, the definition we gave is valid, and yours is too restrictive. However, as we discussed in the study group, computers are rarely so theoretical. There is no purely mathematical set in the Wolfram language, only lists. In the same way, there is no purely theoretical complement implemented, only a computationally approximation of that concept.

Suggestion for Clarifying Section 3 Slide 6

For Section 2 Exercise 3. how do all the paths to (4,2) add up to 7 segments? It looks like 6 to me.

Attachments:

Exercse2_3.nb

Hi!. In the combinatorics, at the start of the course we saw how to count [Not-Replacement +Order Not Relevant] --> Binomial. What would be the approach to count [Replacement + Order Not Relevant], maybe we saw that, but I can't find it. Thanks!

Thanks Marc.. I do believe there may be some impossible (in the sense of not meaningful) cases. I still try to figure out what makes sense to ask and what doesn't .But ok, here I go with a clarification of my question. I still believe this is applicable in real life, but I may be wrong. I tried to depict the situation with an "entry level skills" notebook, so my question gets easier to be assessed..

Thanks Marc, thanks a lot!

Hello Joseph,

Indeed, on problems 1 and 3, there seems to be text from another part of the course. This is an error, as mentionned by Juan Ortiz Navarro, posted 2 days ago Regarding excercises-02.nb The solution to problem 1 (...)

So yes the answer is 25!/15!.

On Excercise 4 of excercises-06.nb, "A smoker is twice as likely to have an ectopic pregnancy as a non-smoking pregnant woman." is denoted as P(E|S)=2P(E). I thought it to be denoted as P(E|S)=2P(E|S'), and P(S|E)=P(E|S)P(S)/(P(E|S)P(S)+P(E|S')P(S'))=0.461538

How can P(E|S)=2P(E|S') be said in words?

Thanks @Juan Ortiz Navarro, I had just calculated the same solution using Bayes theorem. As the solution of "Exercise 4" in the "exercises-06" notebook is different from mine, I consulted this community page.

P( S | E) =2 / (1+1/0.3) = 0.46

Personally, I find the original wording and this solution more interesting..

Mind blowing!...(for me.. for sure there are more advance users for whom this may be already natural.)..but now I can appreciate the flexibility (and why not beauty) of exponential when "connected" to other "devices" to cast a spectrum of other functions. Never expected that to be possible.

Hello Alex,

As noted in the post by John Burke, there is a mistake and the answer is 40, with Binomial[4,3]*Binomial[5,2].

Hi Zbigniew,

The first is also our mistake, now that it is edited. Refer to the original post by John Burke.

The second is a mistake, and will be corrected. Thank you for telling us!

Hello... While reading the second excersise that states "While surfing the web, you encounter ad A 7 times and ads B and C 3 times each. How many arrangements are possible? " I interpreted as having next sequence: AAAAAAABBBCCC or (any other with seven As, three Bs and three Cs, as the precise sequence is not established). So I understood it like n=3 (A,B,C) al possible value reutilization=Yes k (trials)=13 , as the sequence provided, no matter the order is 13 positions long.

So my solution before reading the answer was n^k -> 3^13 possible arrangements. But when reading I found "multinomial". In the question what is the part that suggest that Multinomial approach is the right approach? Thanks

Oh that's interesting, as the assumption for the multinomial is that all possible Ads are "exhausted" at the last observation. My intuition was that, while surfing the internet, I used a finite amount of time, and during that undetermined time Window I saw only 13 Ads, like if I keep browsing I may perfectly see more ads. Let say if I would double the browing time, I may end seeing 26 ads or 30 ads in total, .. So I guess I get your point if all possible ad instances are contained in the limited length sample (of 13 total ads). Otherwise, if not limiting length is stated, I guess it may be fair to assume that As Bs or Cs are never exhausted, and browsing more...means watching more ads..(like the "YouTube Premium thing...that seems to never end.. ;) ....) Thanks Marc for the clarification

Thanks, Marc. I'll keep that advice in mind!

Hello Juan,

This was addressed earlier on this post, but the 0.1 is a mistake. Here is the solution:

P(A'[Intersection]B')=P((A[Union]B)')=1-P(A[Union]B)=1-0.4=0.6

Basically using De Morgan's Law of sets.

Thus, the probability that neither A nor B occurs is 0.6.

1-P(R)=P(R') and 1-P(B)=P(B'). The probability neither happen mean both are false at the same time, thus P(A'[Intersection]B'), a conjunction.

Hopefully that answered your question, although I'm not entirely sure. Let me know if you're still confused.

Regarding excercises-02.nb The solution to problem 1 seems to took a turn I do not understand. Should it be factorial(25)/factorial(10)?

And on problem 3, going from (0,0) to (4,2), are paths of length 7 or 6?

So Binomial[6,2] or Multinomial (4,2)?

Thanks. I meant to write 25!/15! indeed.

Hi J,

This is a bit complicated, so let's discuss this one thing at a time. This textbook expresses a sample space where each data point is in itself a sequence of multiple numbers. In that sense, this is a multivariate random variable. Within a single sample, or data point, there are multiple times, which are the periods of time between breakdowns. Let's say there are n such periods in each sample. Thus, the domain of any single outcome is R(>0)^n, that is, the positive reals in n dimensions (periods of time are always positive).

In this context, the sample space is the set of all possible sequences of periods between breakdowns, possibly R(>0)^n itself.

So overall, I believe you are right with your sample space, but also that you are wrongly interpreting their explaination of the sample space. Hopefully that explanation helped.

Hi Marc, In Lesson 7, Slide 10 and 11, you showed two different examples of how to load and aggregate. I have difficulty in understanding how to aggregate :

height[h_?(60 <= # <= 76 &)] := aggdata[[2, h - 59]]/Length[data] and roll[n_Integer?(2 <= # <= 12 &)] := dice[[n - 1, 2]]/36

Could you elaborate a bit, or use simpler and understandable codes?

Thanks, Lewis

Marc,

You asked for mistakes in the documents. So, please have a look at “Lesson 2, Slide 9”: Your equations are wrong, which means LHS is not equal to RHS. Right?

When is the new study group for Introduction to Probability starting? I missed last week and would rather start from the beginning without worry and rushing.

either of these:

https://wolfr.am/WSG36_Intro_to_Probability or https://amoeba.wolfram.com/index.php/s/LgfWrztXo9xELo4

Marc, I am enjoying the course very much. Thank you for all the effort you are putting into it. When you get time, would you please check Exercise 3 from exercises-04.nb. I did the problem in two different ways, and I still get an answer of 0.6 which does not agree with 0.1 which is your answer. Thanks

Thanks Marc for your prompt consideration. I have another for you! In Exercise 1 of exercises-05.nb, the event you describe out of the 36 possible equally likely outcomes is E={{1, 1}, {2, 2}, {3, 3}, {4, 4}, {5, 5}, {6, 6}, {1, 6}, {6, 1}, {2, 6}, {6, 2}, {3, 6}, {6, 3}, {4, 6}, {6, 4}, {5, 6}, {6, 5}}. Hence, P(E) =16/36=4/9. Thus, the odds are 4 to 5. Would you please look into this also. Thanks.

Marc, I have another one for you. In Exercise 3 of exercises-05.nb, there should be an additional branch to your tree; namely b,b,b,b,r,r. This will give you an additional 1/15 for an answer of 12/15. I found that a simpler way for me to attack the problem was to consider the probability of failure to draw at least one red ball. This means you only have a sub-tree with 3 branches to keep track of. You will then get 3/15 as probability of failure, and hence 12/15 is the probability of success. Thanks for staying on top this.

Marc: I don't quite understand your statement "as soon as I have three balls the game is over". I did assume that once "you" draws a red ball the game is over. I reproduced the terms in your sum but if you carry the tree to the end you get some cases where there are 3 balls left and "you" have not yet drawn a red ball. I work this out in the attached notebook,

Attachments:

Exercises5_Problem3.nb

Marc, Would you check Your answer to Question 8 of the Mock Exam? I get 20*Binomial[19,7] =1,007,760. Thanks.

Hello Willian,

It seems the solution for this problem is wrong and too complicated. To avoid this, let's use the power of the Wolfram language.

Here is the new solution.

Let's compute all arrangements of balls with your friend, where red balls are negative and black balls are positive numbers. We are only interested in the balls that are received by the player, so let's take the odd columns (odd rounds).

possibilities = Permutations[{-1, -2, 1, 2, 3, 4}][[All, {1, 3, 5}]];

Now that we have the balls received in all probabilities, just measure the number of possibilities where we have at least one negative number against the total number of possibilities.

Length@Select[possibilities, AnyTrue[#, Negative] &]/
     Length@possibilities

And we get 4/5. As mentionned by you with the three branches. Surely this way we're less likely to get lost.

Hello all,

We had a more difficult question today, given as:

Given a vector with m element (real numbers) I want to compute the PDF for the sum of the m elements considering that each element will have positive or negative sign with equal probability.

First, consider that each element must have its own probability distribution. Since it is real, let me for example assume any element is uniformly distribution between -10 and 10.

UniformDistribution[{-10, 10}]

If all elements are distributed in the same way, the PDF of the sum will be a result of m elements added. So we get:

PDF[TransformedDistribution[
  m*x, x \[Distributed] UniformDistribution[{-10, 10}]], x]

However, if you assume a normal distribution centered as 0 for your elements, then you could write:

PDF[TransformedDistribution[
  m*x, x \[Distributed] NormalDistribution[0, 1]], x]

Finally, your elements may be distributed in different ways, which could look like something like this, assuming let's say m is 4:

PDF[TransformedDistribution[
  a + b + c + d, {a \[Distributed] NormalDistribution[0, 1], 
   b \[Distributed] NormalDistribution[0, 4], 
   c \[Distributed] NormalDistribution[0, 19], 
   d \[Distributed] UniformDistribution[{-10, 10}]}], x]

Another way to intepret your question is to each element is given a binary choice between -1 or 1, but the values are constant. In that case, you may want to apply many distributions to a single sum, which results in this, assume for example this vector:

v = {1.77, 5.65, 10.14, 195.14}
PDF[TransformedDistribution[
  v[[1]]*(-1)^a + v[[2]]*(-1)^b + v[[3]]*(-1)^c + 
   v[[4]]*(-1)^d, {a, b, c, d} \[Distributed] 
   Table[BernoulliDistribution[0.5], Length[v]]], x]

Hopefully that answers the question. Transformed distributions will be seen in Lesson 20.

Hi,

It may be best for you to prioritize other engagements, as this course is made to be taken at any time.

The course will soon be fully released with all materials and lessons freely accessible. Moreover, the daily study group sessions are all recorded and also freely accessible. There is no planned second study group for this course for now, but it may happen if there is interest for it. Feel free to also continue asking questions in this post even after the study group has ended.

Hi Bob,

I found this proof, from the University of Arizona, which stays within the bounds of the course and you don't need too much external knowledge of other branches of mathematics besides Multivariable Calculus. This is not too much of a difficult proof, but it remains fairly theoretical. The actual proof is short and at the end, but the rest of the paper gives great context.

Hi Bob,

The textbook of the proof is Probability - An Introduction by Geoffrey Grimmett and Dominic Welsh. It is concise but a bit short on examples. A great book for examples would be Introduction to Probability by Charles Grinstead and Laurie Snell. Less mathematical, more examples, more discussion. It's also freely available.

Hopefully that helps.