Talk:High School Mathematics Extensions/Discrete Probability

Ok, to tell you the truth: I do not know much about discrete probabilities. I'm very interested to see how it will develop. I've just seen a A.Level maths exam paper, and the first question is about probability.

P(A)=0.8,P(B)=0.75,P(A"intersection"B)=k.

a)Express P(A"Union"B) in terms of k.

(My)Answer:We sum up P(A) and P(B), but then the probabilities that both A and B are true are counted twice, therefore we subtract k. So it is 0.8+0.75-k=1.55-k

b)Hence show that 0.55&le;k&le;1

(My)Answer:Since all probabilities are smaller than or equal to 1, k must be at least 0.55. Since k itself is also a probability, it must be smaller than or equal to 1. Putting these together, we get 0.55&le;k&le;1

...

--Lemontea 11:49, 3 Feb 2005 (UTC)


 * All correct, and explained very well. To bring your maths to uni level (which you are more than capable of) you need to quote the De Morgan's laws for part a) and actually solve 0 <= 1.55 - k <= 1 algebraicly and use the fact that 0 <= k <=1, for part b). 10/10. Xiaodai 00:25, 6 Feb 2005 (UTC)

I've added some information on adding and multiplying probabilities -- I hope it's all okay. Comments and revisions, of course, are welcome. On another note, I wanted to ask if anyone knows what the final section on options pricing has to do with discrete probability. I personally can't find anything relevant in it.
 * --Buggi22 07:08, 26 May 2005 (UTC)


 * That's okay, and it's an essential part of the topic. However, may I point out that the multiplacation law of probability does not always apply: It is true only when they are independent events. But then, how will we introduce the concept of dependent and independent events, conditional probability etc? Should we put them before or after your section. I'm at a lost at this moment and are looking forward for your work. --Lemontea 09:42, 26 May 2005 (UTC)


 * Option pricing is an important part of finance mathematics. Although it doesn't really use probability in the true sense, it does contain some material on "arbitrage free" probabilities which i think is interesting. Xiaodai 11:08, 27 May 2005 (UTC)

Tree diagram
why there are no three diagrams? 06:14, 22 October 2009 (UTC)

Misconception
I believe there is bias here. I am not expert on the field so I don't dare to edit the page directly. I am talking about this part:

Please note that the probability 1/6 does not mean that it will turn up 1 in at most six tries. Its precise meaning will be discussed later on in the chapter. Roughly, it just means that on the long run (i.e. the die being tossed a large number of times), the proportion of 1's will be very close to 1/6.

There are two school of thought that interpret what does probability means:

http://en.wikipedia.org/wiki/Probability#Interpretations

That misconception just reflect the 1st school of thought, the frequentist while Bayesian school of thought is not properly represented here. 05:56, 22 October 2009 (UTC)05:56, 22 October 2009 (UTC)ArielGenesis (talk) 05:56, 22 October 2009 (UTC)

probability space section feels clunky
I have pasted the following thing here so we can re-use the material later if need be:

We have now all the machinery required to define a Probability Space. Before we do that, let's consider the example of throwing a die again. If we throw a die, the numbers that will turn up are 1, 2, 3, 4, 5 or 6, and nothing else. Notice the events are disjoint and their probabilities add up to 1, i.e. all the possible events have been accounted for.

Definition -- Probability Space
A finite probability space of a random experiment consists of events:
 * $$E_1, E_2,....E_n$$

such that:
 * 1) $$P(E_1) + P(E_2) + ... + P(E_n) = 1$$
 * 2) all events are disjoint

In the above definition a random experiment is simply an action like tossing a coin or picking a card, where the outcome is uncertain. For example, let the random experiment be tossing a die, and let the events be the die turning up 1, 2, 3, 4, 5, or 6. So there are 6 events, and they are all disjoint, and the probabilities of each of the events add up to 1, i.e. all the possible events are included. Therefore it's a probability space.

Example 1

Let the random experiment be choosing a card randomly from a complete deck. And let the events be the card is of the suit Spades, Hearts, Clubs or Diamonds. Is this a probability space?

Solution To check whether it is a probability space, we need to check two conditions. so it's a probability space.
 * 1) There are 4 events, and their probabilities add up to 1.
 * 2) Each of the events are disjoint

Example 2

Let the random experiment be tossing a coin twice. Let the events be
 * the first toss results in a head, or
 * the second toss results in a tail

Is this a probability space?

Solution We proceed to check the two conditions:
 * 1) There are 2 events. Each event has probability of 1/2 (check) and so they add up to 1.
 * 2) The events are not disjoint! So it is NOT a probability space.

Exercises
1. Let the random experiment be tossing a coin once. Let the 2 events be tossing H or T. Is this a probability space

2. Let the random experiment be tossing a die three time. Let the 3 events be the maximum of the 3 die throws are . Is this a probability space
 * 1) 1 or 2,
 * 2) 3 or 4, or
 * 3) 5 or 6

3. Give an exmaple of an experiment that satisfies the first condition but not the second.

4. Give an exmaple of an experiment that satisfies the second condition but not the first.

5. Let S be a probability space. It has n events
 * $$ E_1, E_2, ... E_n$$

Use the Simple Inclusion Exclusion Principle, show that
 * $$ P(E_i \cup E_j) = P(E_i) + P(E_j)$$

for i &ne; j.

6. Is it true that the complement of any event Ei is the union of all the other events i.e. is the following true:
 * $$ \overline{E_i} = E_1 \cup E_2 \cup ... \cup E_i \cup E_{i+1} \cup ... \cup E_n$$

(Hint: use $$ P(\overline{E_1}) = 1 - P(E_1)$$)--Xiaodai 05:57, 6 August 2005 (UTC)

how to pronounce...
Could someone please add some verbiage (in parentheses in the article) that explains how to say/pronounce/express-in-words things like X ~ B.   Is that pronounced "x has a distribution of b"? What about B ~ Bin(n,p)? This would be helpful when trying to converse in person with other mathematicians. Thanks

Impossible and Certain Events equation seems off
Since an answer can only be 0-1, then 0 <= P(E) <= 1 seems wrong, since it would be 0 =>. Greater Than Or Equal To 0, rather than Less Than Or Equal To. As it is, the equation simply says that the event is either 1 or less than 1. Looks like it anyway, am I wrong or is the image?

one addition to help completeness
It's been a while since I studied statistics. I found your presentation easy to follow and purposeful. You used the term (3 over 2) and (n over k). I can't recall what the calculation is for that terminology. And since it is referenced, I don't have an easy way to look it up. Adding a line defining that function would be useful.

Review answer 5
I guess that the question 5 has a mistake on its answer.

As the order doen't matter the P(A=1 and B=3) + P(A=3 and B=1) = 1/6 x 1/6 + 1/6 x 1/6 = 1/18 - Is this now question 6?

''6. Two dice are rolled seperately (sic). What is the probability of getting a 1 and a 3, regardless of order?''

I agree that this seems incorrect, or at least the explanation of the result is inaccurate: the first roll can be either a 1 or a 3. The answer is not as simple as repeating the first half twice: once a result is recorded for the first roll, that result can not be repeated in the second roll; otherwise the result could be two 1s or two 3s. Instead, the probability for the second roll must be 1/6, as it must be either a 1 or a 3, but only whichever of those two was not rolled first. So, if I'm not mistaken, the final probability must be: (1/6 + 1/6) x (1/6) = 2/36 or 1/18

Review answer 10
It looks like answer 10 has mistake. Gareth calculates probability of choosing the student that either plays football or plays video games or studies mathematics. This probability cannot be 11/10 but it cannot be .045 either.

--Deesto (discuss • contribs) 14:54, 31 July 2012 (UTC)

Review Answer 7
I'm a little confused about example number 7:

''Here are the possible combinations: 1 + 6 = 2 + 5 = 3 + 4 = 7"

I think that assumption is correct, but the next one:

"Probability of getting each of the combinations are 1/18 as in exercise 6''

I think it's not:

$$P(A=1) x P(A=6) + P(A=2) x P(A=5) + P(A=3) x P(A=4) = 1/36 + 1/36 + 1/36 = 1/12$$

And the final resolution:

"There are 3 such combinations, so the probability is 3 × 1/18 = 1/6."

Should be:

$$3 x 1/12 = 3 / 12 = 1/4$$

Am I correct?

Thank you all!


 * No. Review the section "Combining addition and multiplication".
 * Instead of


 * $$P(A=1) \times P(A=6)$$ (and similarly for other terms)


 * use


 * $$(P(A=1) \times P(A=6) + P(A=6) \times P(A=1))$$ (and similarly for other terms)


 * so we get


 * $$(1/36 + 1/36) + (1/36 + 1/36) + (1/36 + 1/36) = 2/36 + 2/36 + 2/36 = 3 \times 1/18 = 1/6$$


 * or, alternatively (review Solution 6)


 * $$(P(A=1) + P(A=6)) \times P(opposite)$$


 * so we get


 * $$(2/6 \times 1/6) + (2/6 \times 1/6) + (2/6 \times 1/6) = 2/36 + 2/36 + 2/36 = 3 \times 1/18 = 1/6$$


 * Cartland (discuss • contribs) 05:58, 8 October 2012 (UTC)

Combining addition and multiplication
There is often confusion about the order of multiplying and adding and the resultant meaning. This adds some additional clarification to this section of the main text.

From the main text, we have
 * P(rolling a sum of 3) = P(1 on 1st roll)$$\times$$P(2 on 2nd roll) + P(2 on 1st roll)$$\times$$P(1 on 2nd roll) = $$\frac{1}{6}\times\frac{1}{6}$$ + $$\frac{1}{6}\times\frac{1}{6}$$ = 1/18 $$\approx$$ 0.056 (or 5.6%)

In what way is this different from
 * P(?) = (P(1 on 1st roll) + P(2 on 2nd roll)) $$\times$$ (P(2 on 1st roll) + P(1 on 2nd roll)) = ($$\frac{1}{6} + \frac{1}{6}$$) $$\times$$ ($$\frac{1}{6} + \frac{1}{6}$$) = $$\frac{2}{6}\times\frac{2}{6}$$ = 1/9 $$\approx$$ 0.111 (or 11.1%) ?

The short answer is because it's answering a different question: not P(rolling a sum of 3) but P(only 1 or 2 in both rolls).

Sample spaces can be useful to see what's going on (https://en.wikipedia.org/wiki/Probability_space).

For simplicity, consider a 4 sided dice (pyramid). The sample space of two rolls (or 2 dice rolled together), represented by single digit decimals, is (16 possibilities):

(Hopefully it is obvious that, for example, 23 is not twenty three, but represents 2 on one roll and 3 on the other roll ;).

P(rolling a sum of 3) contains 12 and 21

= P(1 on 1st roll)$$\times$$P(2 on 2nd roll) + P(2 on 1st roll)$$\times$$P(1 on 2nd roll) = $$\frac{1}{4}\times\frac{1}{4}$$ + $$\frac{1}{4}\times\frac{1}{4}$$ = $$\frac{1}{16} + \frac{1}{16}$$ = 2/16 = 1/8

P(only 1 or 2 in both rolls) contains 11 and 12 and 21 and 22

= (P(1 on 1st roll) + P(2 on 2nd roll)) $$\times$$ (P(2 on 1st roll) + P(1 on 2nd roll)) = ($$\frac{1}{4} + \frac{1}{4}$$) $$\times$$ ($$\frac{1}{4} + \frac{1}{4}$$) = $$\frac{2}{4}\times\frac{2}{4}$$ = 4/16 = 1/4

--Cartland (discuss • contribs) 07:02, 8 October 2012 (UTC)

No Coverage of Conditional Probability
I think this page is a useful starting point. Some points for improvement (and apologies if I have mistakes of notation or process - I'm new to this form of communication):


 * There are vague references at the end to distinctly non-discrete topics like the uniform distribution. Placeholders I could understand, but not to areas that are off topic for the this page.
 * Some coverage of conditional probablility would be useful. There are some simple concepts and notations that can be explained with the kind of venn diagram samples that are nicely used in parts of the existing material, e.g. Bayes Theorem.
 * I echo someone else's comment that (3 over 2) is a notation that needs to be introduced if it is going to be used. Better perhaps not to use it at that point in the text.

Please explain the notation
In the section The Bernoulli experiment we have the notation:


 * $$B \sim Ber(p) $$

what does the function $$ Ber(p) $$ mean? I am not sure what this section is saying, but I guess it is the probability of result p? -- Q Chris (discuss • contribs) 13:52, 8 November 2012 (UTC)

Venn diagram claim
I am confused by this claim in the Venn diagram section: "It is generally very difficult to draw Venn diagrams for more than 3 intersecting sets. To demonstrate why, here is a Venn diagram showing four intersecting sets." Essentially, this says "here is a diagram of something very difficult to draw", and yet there it is -- a fairly simple diagram. Am I misinterpreting this? --Ds13 (discuss • contribs) 23:10, 18 November 2012 (UTC)

It's rather hard to find intersection of all 4 sets. Try yourself. --ViolentOr (discuss • contribs) 13:42, 20 January 2013 (UTC)

Exercise 9
I think all is well explained and there is no too much content, but perhaps the proposed solution for the exercise 9 is a little tricky for some students. A more consistent solution with the exposed explanations could be:

P(C>A) = P([(C=1)⋂(A>1)] ⋃ [(C=2)⋂(A>2)] ⋃ [(C=3)⋂(A>3)] ⋃ [(C=4)⋂(A>4)] ⋃ [(C=5)⋂(A>5)]) ; Note: P(A>6) = 0;

P(C>A) = P(C=1).P(A>1)+P(C=2).P(A>2)+P(C=3).P(A>3)+P(C=4).P(A>4)+P(C=5).P(A>5);

Since P(C=1) = P(C=2) = P(C=3) = P(C=4) = P(C=5) = 1/6 = P(C=any) for 1 ≤ any ≤ 6 :

P(C>A) = P(C=any).(P(A>1)+P(A>2)+P(A>3)+P(A>4)+P(A>5));

And since P(A>i) = (6-i)/6 for 1 ≤ i ≤ 6 :

P(C>A) = 1/6.(5/6 + 4/6 + 3/6 + 2/6 + 1/6) = 15/36 = 5/12

A discussion of the Monte Hall "paradox" (from the gameshow) is always a good practical example.
http://en.wikipedia.org/wiki/Monty_Hall_problem

Optional exercise - ex. 7
Is this the only optional one? If it is the only optional exercise, wouldn't it make sense to move it to the end of the exercises. Otherwise it seems weird to have one exercise that is unecessary then more to do.

If ex.8 - ex.10 are also optional, they need marking as such.