# Probability question: Independant events **URL:** \ **Category:** Factual Questions\ **Created:** [February 3, 2012, 2:22pm UTC](https://boards.straightdope.com/t/probability-question-independant-events/611649 "2012-02-03T14:22:50Z")\ **Posts on this page:** 10\ **Page:** 1
**Author:** ![Saffer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saffer/32/6440_2.png) [@Saffer](https://boards.straightdope.com/u/Saffer)\ **Post date:** [February 3, 2012, 2:22pm UTC](https://boards.straightdope.com/t/probability-question-independant-events/611649/1 "2012-02-03T14:22:50Z")
I have been reading this book: > **[Measure, Integral and Probability (Springer Undergraduate Mathematics Series):...](https://www.amazon.co.uk/Integral-Probability-Springer-Undergraduate-Mathematics/dp/1852337818)** > > Buy Measure, Integral and Probability (Springer Undergraduate Mathematics Series) Softcover reprint of the original 2nd ed. 2004 by Capinski, Marek, Kopp, Peter E. (ISBN: 9781852337810) from Amazon's Book Store. Everyday low prices and free delivery... On page 48, the book says: “Let Ω = [0,1], A = [0,1/4] as before; then A is independant of B = [1/8,5/8] and of C = [1/8,3/8] U [3/4,1]. In addidtion, B and C are independant. However, P(A∩B∩C) ≠ P(A)P(B)P(C). Thus, given three events, the pairwise independance of each of the three possible pairs does not sufice for the extension of ‘independance’ to all three events.” When I first read this I was like :dubious: so I put together a spreadsheet using the RAND function to get over 100000 rows of random numbers in columns A,B and C. In column D I put a big IF statment that returns a 1 if event A above occurs in column A and event B occurs in column B etc. It returns 0 if even one of the events do not occur. On their own, I calculate the probability of event A = 0.25, P(B) = 0.5, P(C) = 0.5 Then P(A)P(B)P(C) = 0.0625 If I do a sum of column D on my spreadsheet and divide that by a count then it gets quite close to 0.0625. Obviously never exacly the same but never too far. So what is the author on about? ---
**Author:** ![John\_DiFool](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_difool/32/19543_2.png) [@John\_DiFool](https://boards.straightdope.com/u/John_DiFool)\ **Post date:** [February 3, 2012, 2:41pm UTC](https://boards.straightdope.com/t/probability-question-independant-events/611649/2 "2012-02-03T14:41:39Z")
Independent events should always be multiplied together, and .0625 is what you get from (0.5)_(0.5)_(0.25). ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [February 3, 2012, 2:56pm UTC](https://boards.straightdope.com/t/probability-question-independant-events/611649/3 "2012-02-03T14:56:36Z")
The problem is that A∩B∩C = A∩B = A∩C – it’s the interval [1/8,1/4], and so has probability 0.125. You don’t need a spreadsheet with random functions for that: you just need to draw a diagram, and see on that where sets A, B and C intersect. ---
**Author:** ![Saffer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saffer/32/6440_2.png) [@Saffer](https://boards.straightdope.com/u/Saffer)\ **Post date:** [February 3, 2012, 2:57pm UTC](https://boards.straightdope.com/t/probability-question-independant-events/611649/4 "2012-02-03T14:57:45Z")
> [@John\_DiFool](#): > > Independent events should always be multiplied together, and .0625 is what you get from (0.5)_(0.5)_(0.25). So are you saying, based on what I have quoted above, that the book is no good? ---
**Author:** ![Saffer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saffer/32/6440_2.png) [@Saffer](https://boards.straightdope.com/u/Saffer)\ **Post date:** [February 3, 2012, 3:05pm UTC](https://boards.straightdope.com/t/probability-question-independant-events/611649/5 "2012-02-03T15:05:22Z")
> [@Giles](#): > > The problem is that A∩B∩C = A∩B = A∩C – it’s the interval [1/8,1/4], and so has probability 0.125. You don’t need a spreadsheet with random functions for that: you just need to draw a diagram, and see on that where sets A, B and C intersect. I don’t think that approach is appropriate for independant events. What you are saying is as if I only had random numbers in one column and was testing for each condition (A,B,C) on this column. As these are all independant, each condition is tested on a sepparate column of random numbers. So the result is different. No? ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [February 3, 2012, 3:15pm UTC](https://boards.straightdope.com/t/probability-question-independant-events/611649/6 "2012-02-03T15:15:11Z")
If you are taking random numbers in the interval [0,1], then: A is satisfied if the number is between 0 and 1/4 B is satisfied if the number is between 1/8 and 5/8 C is satisfied if the number is between 1/8 and 3/8, or between 3/4 and 1 So all three of A, B and C are satisfied if the number is between 1/8 and 1/4 – but each time you should be checking the same number. Is that how your spreadsheet works? ---
**Author:** ![Saffer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saffer/32/6440_2.png) [@Saffer](https://boards.straightdope.com/u/Saffer)\ **Post date:** [February 3, 2012, 3:20pm UTC](https://boards.straightdope.com/t/probability-question-independant-events/611649/7 "2012-02-03T15:20:06Z")
> [@Giles](#): > > If you are taking random numbers in the interval [0,1], then: > > A is satisfied if the number is between 0 and 1/4 > B is satisfied if the number is between 1/8 and 5/8 > C is satisfied if the number is between 1/8 and 3/8, or between 3/4 and 1 > > So all three of A, B and C are satisfied if the number is between 1/8 and 1/4 – but each time you should be checking the same number. > > Is that how your spreadsheet works? It is more like this: A is satisfied if the number1 is between 0 and 1/4 B is satisfied if the number2 is between 1/8 and 5/8 C is satisfied if the number3 is between 1/8 and 3/8, or between 3/4 and 1 Also, I have just realised the point the author was making: The probability of the intersection does not = the product of the probabilities for independant events. So you calculaiton of 0.125 does not = 0.0625 ---
**Author:** ![Buck\_Godot](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/buck_godot/32/6573_2.png) [@Buck\_Godot](https://boards.straightdope.com/u/Buck_Godot)\ **Post date:** [February 3, 2012, 5:02pm UTC](https://boards.straightdope.com/t/probability-question-independant-events/611649/8 "2012-02-03T17:02:02Z")
I still think your missing the main point which is that even though any pair of these events is independent, the combination of all three isn’t. From the example P(A)=1/4 (area between 0 and 1/4) P(B)=1/2 (area between 1/8 and 5/8) P©=1/2 (area between 1/8 and 3/8, or between 3/4 and 1) The probability of the intersection P(A and B) = 1/8 (area between 1/8 and 1/4) = P(A)\*P(B) so (A and B are independent) P(A and C) = 1/8 (area between 1/8 and 1/4) = P(A)\*P© so (A and C are independent) P(B and C) = 1/4 (area between 1/8 and 3/8) = P(B)\*P© so (B and C are independent) However P(A and B and C) = 1/8 (area between 1/8 and 1/4) While P(A)\*P(B)\*P© = 1/16 so the triplet (A,B,C) is not independent. By themselves knowing I’m in region A tells me nothing about whether I’m in region B (I have a 1/4 chance whether B is true or not) and vis-versa P(B)=P(B|A)=P(B|!A)=1/2 Similarly A and C provide no information about each other, and B and C provide no information about each other. However in combination, if I know that B and C are both true, then that tells me that A must be true. So even though every pair is independent, the whole set of events isn’t. Here is simpler example: I flip 2 coins and define the following three events. A = first coin is heads B = second coin is heads C = both coins show the same face. It is clear that A and B are independent, since there is no relationship. Also I just know A that the first coin is a head, I still don’t have any information about whether the 2 coins have the same face, so A and C are independent. Similarly B and C are independent. However knowledge of both A and B in combination gives me complete knowledge of C. So the combined three events aren’t independent. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [February 3, 2012, 6:44pm UTC](https://boards.straightdope.com/t/probability-question-independant-events/611649/9 "2012-02-03T18:44:50Z")
Another example: Roll a four-sided die. Let A = {1, 2}, B = {2, 4} and C = {2, 3}. P(A and B) = 1/4 = P(A)P(B) P(A and C) = 1/4 = P(A)P© P(B and C) = 1/4 = P(B)P© However, P(A and B and C) = 1/4, but P(A)P(B)P© = 1/8. So the events A, B and C are _pairwise independent_, but not independent. ---
**Author:** ![Saffer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saffer/32/6440_2.png) [@Saffer](https://boards.straightdope.com/u/Saffer)\ **Post date:** [February 4, 2012, 3:10pm UTC](https://boards.straightdope.com/t/probability-question-independant-events/611649/10 "2012-02-04T15:10:52Z")
> [@](#): > > I still think your missing the main point… You were right, I was missing that point :smack: Thanks to Buck Godot and ultrafilter for clearing that up.