##  General Advance-Placement (AP) Statistics Curriculum \- Probability Theory Rules

###  Addition Rule

The probability of a union, also called the Inclusion-Exclusion principle allows us to compute probabilities of composite events represented as unions (i.e., sums) of simpler events. 

Venn Diagrams

For events _A_ 1, ..., _A_ _n_ in a probability space (S,P), the probability of the union for _n=2_ is 

    

For _n=3_ , 

    

In general, for any _n_ , 

    
    

###  Conditional Probability

The conditional probability of A occurring given that B occurs is given by 

    

  * See this  demonstration of conditional probability using the SOCR Bivariate Normal Experiment. 


  * Also see  demonstration of conditional probability using the SOCR Coin-Die Experiment. 



###  Examples

####  Contingency table

Here is the data on 400 Melanoma (skin cancer) Patients by Type and Site 

Type | Site | Totals   
---|---|---  
Head and Neck |  Trunk |  Extremities   
Hutchinson's melanomic freckle |  22 |  2 |  10 |  34   
Superficial |  16 |  54 |  115 |  185   
Nodular |  19 |  33 |  73 |  125   
Indeterminant |  11 |  17 |  28 |  56   
Column Totals |  68 |  106 |  226 |  400   
  
  * Suppose we select one out of the 400 patients in the study and we want to find the probability that the cancer is on the extremities _given_ that it is of type nodular: P = 73/125 = P(Extremities | Nodular) 


  * What is the probability that for a randomly chosen patient the cancer type is Superficial given that it appears on the Trunk? 



####  Monty Hall Problem

Recall that earlier we discussed the  Monty Hall Experiment. We will now show why the odds of winning double if we use the swap strategy - that is the probability of a win is 2/3, if each time we switch and choose the last third card. 

Denote W={Final Win of the Car Price}. Let L1 and W2 represent the events of choosing the donkey (loosing) and the car (winning) at the player's first and second choice, respectively. Then, the chance of winning in the swapping-strategy case is: . If we played using the  stay-home strategy, our chance of winning would have been: , or half the chance in the first (swapping) case. 

####  Drawing balls without replacement

Suppose we draw 2 balls at random, one at a time without replacement from an urn containing 4 black and 3 white balls, otherwise identical. What is the probability that the second ball is black? Sample Space? P({2-nd ball is black}) = P({2-nd is black} &{1-st is black}) + P({2-nd is black} &{1-st is white}) = 4/7 x 3/6 + 4/6 x 3/7 = 4/7. 

###  Inverting the order of conditioning

In many practical situations it is beneficial to be able to swap the event of interest and the conditioning event when we are computing probabilities. This can easily be accomplished using this trivial, yet powerful, identity: 

###  Example - inverting conditioning

Suppose we classify the entire female population into 2 Classes: healthy(NC) controls and cancer patients. If a woman has a positive mammogram result, what is the probability that she has breast cancer? 

Suppose we obtain medical evidence for a subject in terms of the results of her mammogram (imaging) test: positive or negative mammogram . If P(Positive Test) = 0.107, P(Cancer) = 0.1, P(Positive test | Cancer) = 0.8, then we can easily calculate the probability of real interest - what is the chance that the subject has cancer: 

This equation has 3 known parameters and 1 unknown variable, so, we can solve for P(Cancer | Positive Test) to determine the chance the patient has breast cancer given that her mammogram was positively read. This probability, of course, will significantly influence the treatment action recommended by the physician. 

###  Statistical Independence

Events A and B are **statistically independent** if knowing whether B has occurred gives no new information about the chances of A occurring, i.e., if P(A | B) = P(A). 

Note that if A is independent of B, then B is also independent of A, i.e., P(B | A) = P(B), since . 

If A and B are statistically independent, then 

###  Multiplication Rule

For any two events (whether dependent or independent): 

In general, for any collection of events: 

###  Law of total probability

If {} partition the sample space _S_ (i.e., all events are mutually exclusive and ) then for any event B 

  * Example, if _A_ 1 and _A_ 2 partition the sample space (think of males and females), then the probability of any event B (e.g., smoker) may be computed by: 



_P_(_B_) = _P_(_B_ | _A_ 1)_P_(_A_ 1) + _P_(_B_ | _A_ 2)_P_(_A_ 2). This of course is a simple consequence of the fact that . Therefore, . 

###  Bayesian Rule

If {} partition the sample space _S_ and A and B are any events (subsets of S), then: 

###  Independence vs. disjointness/mutual-exclusiveness

     The events A and B are _independent_ if P(A|B)=P(A). That is 
     The events C and D are _disjoint, or mutually-exclusive_ , if . That is 

Mutual-exclusiveness and independence are different concepts. Here are two examples clarifying the differences between these concepts: 

  * Suppose we play a  card game of _guessing the color of a randomly drawn card_ from a standard 52-card deck. As there are 2 possible colors (black and red), and given no other information, the chance for correctly guessing the color (e.g., black) is 0.5. However, additional information may or may not be helpful in identifying the card color. For example: 
    * If we know that the card denomination is a king, there are 2 red and 2 black kings, this does **not** help us improve our chances of successfully identifying the correct color of the card, P(Red|King)=P(Red), independence. 
    * If we know that the suit of the card is hearts, this does help is with correctly identifying the card color (as hearts are red), P(Red|Hearts)=1.0, strong dependence. 
    * Notes: 



    

     In both cases, the events A={Red} and B={King} and C={Hearts} are **not** mutually exclusive (disjoint)! 
     Events that are mutually exclusive (disjoint) cannot be independent! 

  * Color blindness is a sex-linked trait, as many of the genes involved in color vision are on the X chromosome. Color blindness more common in males than in females, as men do not have a second X chromosome to overwrite the chromosome which carries the mutation. If 8% of variants of a given gene are defective (mutated), the probability of a single copy being defective is 8%, but the probability that two (independent) copies are both defective is 0.08 × 0.08 = 0.0064. 



     The events A={Female} and B={Color blind} are not mutually exclusive (females can be color blind), nor they are independent (the rate of color blindness among females is lower). Color blindness prevalence within the 2 genders is P(CB|Male) = 0.08, and P(CB|Female)=0.005, where CB={color blind, one color, a color combination, or another mutation}. 

###  Example

Suppose a Laboratory blood test is used as evidence for a disease. Assume P(positive Test| Disease) = 0.95, P(positive Test| no Disease)=0.01 and P(Disease) = 0.005. Find P(Disease|positive Test)=? 

Denote D = {the test person has the disease}, _D_ _c_ = {the test person does not have the disease} and T = {the test result is positive}. Then