AP Statistics Curriculum 2007 Prob Rules
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<center><math>P(D  T) = {P(T  D) P(D) \over P(T)} = {P(T  D) P(D) \over P(TD)P(D) + P(TD^c)P(D^c)}= {0.95\times 0.005 \over {0.95\times 0.005 +0.01\times 0.995}}=0.193.</math></center>  <center><math>P(D  T) = {P(T  D) P(D) \over P(T)} = {P(T  D) P(D) \over P(TD)P(D) + P(TD^c)P(D^c)}= {0.95\times 0.005 \over {0.95\times 0.005 +0.01\times 0.995}}=0.193.</math></center>  
  ==See also==  +  ===See also=== 
* [[AP_Statistics_Curriculum_2007_Bayesian_Prelim  Bayesian Chapter]]  * [[AP_Statistics_Curriculum_2007_Bayesian_Prelim  Bayesian Chapter]]  
Revision as of 22:21, 22 October 2009
Contents

General AdvancePlacement (AP) Statistics Curriculum  Probability Theory Rules
Addition Rule
The probability of a union, also called the InclusionExclusion principle allows us to compute probabilities of composite events represented as unions (i.e., sums) of simpler events.
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
Examples
Contingency table
Here is the data on 400 Melanoma (skin cancer) Patients by Type and Site
Site  
Type  Head and Neck  Trunk  Extremities  Totals 
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 L_{1} and W_{2} 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 swappingstrategy case is: . If we played using the stayhome 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({2nd ball is black}) = P({2nd is black} &{1st is black}) + P({2nd is black} &{1st is white}) = 4/7 x 3/6 + 4/6 x 3/7 = 4/7.
Inverting the order of conditioning
In many practical situations is 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 {} form a partition of 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 .
Bayesian Rule
If {} form a partition of the sample space S and A and B are any events (subsets of S), then:
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(Diseasepositive 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
See also
References
 SOCR Home page: http://www.socr.ucla.edu
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