Successfully reported this slideshow.

×

# UNIT-II-Probability-ConditionalProbability-BayesTherom.pptx

Probability

Probability

## More Related Content

### UNIT-II-Probability-ConditionalProbability-BayesTherom.pptx

1. 1. Probability
2. 2. • Example: Soccer Game • You are off to soccer, and love being the Goalkeeper, but that • depends who is the Coach today: • with Coach Sam the probability of being Goalkeeper is 0.5 • with Coach Alex the probability of being Goalkeeper is 0.3 • Sam is Coach more often ... about 6 out of every 10 games (a probability of 0.6). • So, what is the probability you will be a Goalkeeper today? • Let's build the tree diagram. First we show the two possible coaches: Sam or Alex:
3. 3. • The probability of getting Sam is 0.6, so the probability of Alex must be 0.4 (together the • probability is 1) • Now, if you get Sam, there is 0.5 probability of being Goalie (and 0.5 of not being Goalie): If you get Alex, there is 0.3 probability of being Goalie (and 0.7 not):
4. 4. • (When we take the 0.6 chance of Sam being coach and include the 0.5 chance that Sam will • let you be Goalkeeper we end up with an 0.3 chance.) • But we are not done yet! We haven't included Alex as Coach: • An 0.4 chance of Alex as Coach, followed by an 0.3 chance gives 0.12. • Now we add the column: • 0.3 + 0.12 = 0.42 probability of being a Goalkeeper today • (That is a 42% chance)
5. 5. Conditional Probability
6. 6. Fundamentals for Bayes Theorem ● 0 ≤ P(A) ≤ 1, where P(A) is the probability of an event A. ● P(A) = 0, indicates total uncertainty in an event A. ● P(A) =1, indicates total certainty in an event A. ● Event: Each possible outcome of a variable is called an event. ● Sample space: The collection of all possible events is called sample space. ● Random variables: Random variables are used to represent the events and objects in the real world. ● Prior probability: The prior probability of an event is probability computed before observing new information. ● Posterior Probability: The probability that is calculated after all evidence or information has taken into account. It is a combination of prior probability and new information.
7. 7. ● Bayes' theorem is also known as Bayes' rule, Bayes' law, or Bayesian reasoning, which determines the probability of an event with uncertain knowledge. ● In probability theory, it relates the conditional probability and marginal probabilities of two random events. ● Bayes' theorem was named after the British mathematician Thomas Bayes. The Bayesian inference is an application of Bayes' theorem, which is fundamental to Bayesian statistics. ● It is a way to calculate the value of P(B|A) with the knowledge of P(A|B).
8. 8. • Bayes' Theorem is a way of finding a probability when we know certain other probabilities. The formula is: • P(A|B) = P(A) P(B|A)/P(B) • Which tells us: how often A happens given that B happens, written P(A|B), • When we know: how often B happens given that A happens, written P(B|A) • and how likely A is on its own, written P(A) • and how likely B is on its own, written P(B)
9. 9.  Bayes's theorem is expressed mathematically by the following equation that is given below.  𝐏 𝐗 ∣ 𝐘 = 𝐏 𝐘∣𝐗 𝐏 𝐗 𝐏 𝐘  Where X and Y are the events and P (Y) ≠ 0  P(X/Y) is a conditional probability that describes the occurrence of event X is given that Y is true.  P(Y/X) is a conditional probability that describes the occurrence of event Y is given that X is true.  P(X) and P(Y) are the probabilities of observing X and Y independently of each other. This is known as the marginal probability.  Let us say P(Fire) means how often there is fire, and P(Smoke) means how often we see smoke, then:  P(Fire|Smoke) means how often there is fire when we can see smoke  P(Smoke|Fire) means how often we can see smoke when there is fire  So the formula kind of tells us "forwards" P(Fire|Smoke) when we know "backwards" P(Smoke|Fire)
10. 10.  Example:  dangerous fires are rare (1%)  but smoke is fairly common (10%) due to barbecues,  and 90% of dangerous fires make smoke  We can then discover the probability of dangerous Fire when there is Smoke:  P(Fire|Smoke) = P(Fire) P(Smoke|Fire)/P(Smoke)  = 1% x 90%10%  = 9%
11. 11. ● Bayes' theorem can be derived using product rule and conditional probability of event A with known event B: ● As from product rule we can write: P(A ⋀ B)= P(A|B) P(B) or ● Similarly, the probability of event B with known event A: P(A ⋀ B)= P(B|A) P(A) ● According to above statements the equation for Bayes’ rule or Bayes’ theorem is as below:  𝐏 𝐀 ∣ 𝐁 = 𝐏 𝐁∣𝐀 𝐏 𝐀 𝐏 𝐁 ● P(A|B) is known as posterior, which we need to calculate, and it will be read as Probability of hypothesis A when we have occurred an evidence B. ● P(B|A) is called the likelihood, in which we consider that hypothesis is true, then we calculate the probability of evidence. ● P(A) is called the prior probability, probability of hypothesis before considering the evidence ● P(B) is called marginal probability, pure probability of an evidence. ● In the equation (a), in general, we can write P (B) = P(A)*P(B|Ai), hence the Bayes' rule can be written as:  𝐏 𝐀𝐢 ∣ 𝐁 = 𝐏 𝐀𝐢 𝐏 𝐁∣𝐀𝐢 𝐢=𝟏 𝐊 𝐏 𝐀𝐢 ∗𝐏 𝐁∣𝐀𝐢  Where A1, A2, A3, ..............., An is a set of mutually exclusive and exhaustive events.
12. 12.  Example: In a particular pain clinic, 10% of patients are prescribed narcotic pain killers. Overall, 5% of the clinic's patients are addicted to narcotics(including pain killers and illegal substances). Out of all the people restricted pain pills, 8% are addicts. If a patient is an addict, what is the probability that they will be prescribed pain pills?  A: Being prescribed pain pills 10%  B: Being an addict 5%  B|A: Out of all 8% addict for narcotics.  P(A|B)=P(B|A)P(A)/P(B)  P(A|B)=(0.08 * 0.1)/0.05 =0.16  That means  If a patient is an addict, then there are 16% chances to prescribe pain pills.
13. 13. ● According to the given statement the below are statistics: ● The prior probability of having cancer among population is P(Cancer) Which is P(Cancer)=0.08 ● The prior probability of not having cancer among population is P(‫ך‬Cancer) Which is P(‫ך‬Cancer)=1-0.08=0.992 ● The cancer test results are accurate up to 98%. It is indicated as P(+|Cancer)=0.98 and 2% of results are negative such as: P(‫|ך‬Cancer)=0.2(1-0.98) ● Probability of positive result for no cancer is P(+|‫ך‬Cancer)=1-0.97=0.3 ● Probability of negative result for no cancer is P(-|‫ך‬Cancer)=0.97
14. 14. Bayes’ Rule Let S1 , S2 , S3 ,..., Sk be mutually exclusive and exhaustive events with prior probabilities P(S1), P(S2),…,P(Sk). If an event A occurs, the posterior probability of Si, given that A occurred is ,...k , i S A P S P S A P S P A S P i i i i i 2 1 for ) | ( ) ( ) | ( ) ( ) | (    ) | ( ) ( ) | ( ) ( ) ( ) ( ) | ( ) | ( ) ( ) ( ) ( ) ( ) | ( Proof i i i i i i i i i i i i S A P S P S A P S P A P AS P A S P S A P S P AS P S P AS P S A P       
15. 15. We know: P(F) = P(M) = P(H|F) = P(H|M) = Example From a previous example, we know that 49% of the population are female. Of the female patients, 8% are high risk for heart attack, while 12% of the male patients are high risk. A single person is selected at random and found to be high risk. What is the probability that it is a male? Define H: high risk F: female M: male 61 . ) 08 (. 49 . ) 12 (. 51 . ) 12 (. 51 . ) | ( ) ( ) | ( ) ( ) | ( ) ( ) | (      F H P F P M H P M P M H P M P H M P .12 .08 .51 .49
16. 16. Example Suppose a rare disease infects one out of every 1000 people in a population. And suppose that there is a good, but not perfect, test for this disease: if a person has the disease, the test comes back positive 99% of the time. On the other hand, the test also produces some false positives: 2% of uninfected people are also test positive. And someone just tested positive. What are his chances of having this disease?
17. 17. We know: P(A) = .001 P(Ac) =.999 P(B|A) = .99 P(B|Ac) =.02 Example Define A: has the disease B: test positive 0472 . 02 . 999 . 99 . 001 . 99 . 001 . ) | ( ) ( ) | ( ) ( ) | ( ) ( ) | (         c A B P c A P A B P A P A B P A P B A P We want to know P(A|B)=?
18. 18. Example A survey of job satisfaction2 of teachers was taken, giving the following results 2 “Psychology of the Scientist: Work Related Attitudes of U.S. Scientists” (Psychological Reports (1991): 443 – 450). Satisfied Unsatisfied Total College 74 43 117 High School 224 171 395 Elementary 126 140 266 Total 424 354 778 Job Satisfaction L E V E L
19. 19. Example If all the cells are divided by the total number surveyed, 778, the resulting table is a table of empirically derived probabilities. Satisfied Unsatisfied Total College 0.095 0.055 0.150 High School 0.288 0.220 0.508 Elementary 0.162 0.180 0.342 Total 0.545 0.455 1.000 L E V E L Job Satisfaction
20. 20. Example For convenience, let C stand for the event that the teacher teaches college, S stand for the teacher being satisfied and so on. Let’s look at some probabilities and what they mean. is the proportion of teachers who are college teachers. P(C) 0.150  is the proportion of teachers who are satisfied with their job. P(S) 0.545  is the proportion of teachers who are college teachers and who are satisfied with their job. P(C S) 0.095  Satisfied Unsatisfied Total College 0.095 0.055 0.150 High School 0.288 0.220 0.508 Elementary 0.162 0.180 0.342 Total 0.545 0.455 1.000 L E V E L Job Satisfaction
21. 21. Example is the proportion of teachers who are college teachers given they are satisfied. Restated: This is the proportion of satisfied that are college teachers. P(C S) P(C | S) P(S) 0.095 0.175 0.545    is the proportion of teachers who are satisfied given they are college teachers. Restated: This is the proportion of college teachers that are satisfied. P(S C) P(S | C) P(C) P(C S) 0.095 P(C) 0.150 0.632     Satisfied Unsatisfied Total College 0.095 0.055 0.150 High School 0.288 0.220 0.508 Elementary 0.162 0.180 0.342 Total 0.545 0.455 1.000 L E V E L Job Satisfaction
22. 22. Example P(C S) 0.095 P(C) 0.150 and P(C | S) 0.175 P(S) 0.545     Satisfied Unsatisfied Total College 0.095 0.055 0.150 High School 0.288 0.220 0.508 Elementary 0.162 0.180 0.342 Total 0.545 0.455 1.000 L E V E L Job Satisfaction P(C|S)  P(C) so C and S are dependent events. Are C and S independent events?
23. 23. Example Satisfied Unsatisfied Total College 0.095 0.055 0.150 High School 0.288 0.220 0.508 Elementary 0.162 0.180 0.658 Total 0.545 0.455 1.000 L E V E L Job Satisfaction P(C) = 0.150, P(S) = 0.545 and P(CS) = 0.095, so P(CS) = P(C)+P(S) - P(CS) = 0.150 + 0.545 - 0.095 = 0.600 P(CS)?
24. 24. Tom and Dick are going to take a driver's test at the nearest DMV office. Tom estimates that his chances to pass the test are 70% and Dick estimates his as 80%. Tom and Dick take their tests independently. Define D = {Dick passes the driving test} T = {Tom passes the driving test} T and D are independent. P (T) = 0.7, P (D) = 0.8 Example
25. 25. What is the probability that at most one of the two friends will pass the test? Example P(At most one person pass) = P(Dc  Tc) + P(Dc  T) + P(D  Tc) = (1 - 0.8) (1 – 0.7) + (0.7) (1 – 0.8) + (0.8) (1 – 0.7) = .44 P(At most one person pass) = 1-P(both pass) = 1- 0.8 x 0.7 = .44
26. 26. What is the probability that at least one of the two friends will pass the test? Example P(At least one person pass) = P(D  T) = 0.8 + 0.7 - 0.8 x 0.7 = .94 P(At least one person pass) = 1-P(neither passes) = 1- (1-0.8) x (1-0.7) = .94
27. 27. Suppose we know that only one of the two friends passed the test. What is the probability that it was Dick? Example P(D | exactly one person passed) = P(D  exactly one person passed) / P(exactly one person passed) = P(D  Tc) / (P(D  Tc) + P(Dc  T) ) = 0.8 x (1-0.7)/(0.8 x (1-0.7)+(1-.8) x 0.7) = .63