1. BIOSTAT 3 Three tradition views of probabilities: Classical approach: make certain assumptions (such as equally likely, independence) about situation. [roll die P(2) = 1/6 = 0.167] Relative frequency: assigning probabilities based on experimentation or historical data. [light 100 firecrackers, 80 pop, P(firecracker pops) = 80/100 = 0.8] Subjective [Bayesian] approach: Assigning probabilities based on the assignor’s judgment. [Based on my past teaching experience I believe that P(a student in this class passes) = 0.98] As a student, would you change this probability if you made a 15 on the midterm exam?

2. BIOSTAT 3 Given k possible (mutually exclusive and collectively exhaustive) outcomes to an experiment {O 1, O 2, …, O k }, the probabilities assigned to the outcomes must satisfy these requirements: (1)The probability of any outcome is between 0 and 1 i.e. 0 ≤ P(O i ) ≤ 1 for each i, and (2)The sum of the probabilities of all the outcomes equals 1 i.e. P(O 1 ) + P(O 2 ) + … + P(O k ) = 1

3. BIOSTAT 3 There are several types of combinations and relationships between events: Complement of an event [everything other than that event] Intersection of two events [event A and event B] or [A*B] Union of two events [event A or event B] or [A+B] Conditional probability P(A/B): read the probability of A given B has occurred.

4. BIOSTAT 3 Conditional Probability Joint Probability (Multiplication rule) Additional rule

5. BIOSTAT 3 Independent events: the probability of one event is not affected by the occurrence of the other event. Two events A and B are said to be independent if P(A|B) = P(A) and P(B|A) = P(B) P(you have a flat tire going home/radio quits working)

6. BIOSTAT 3 A and B are mutually exclusive if the occurrence of one event makes the other one impossible. This means that P(A and B) = P(A * B) = 0 The addition rule for mutually exclusive events is P(A or B) = P(A) + P(B) Only if A and B are Mutually Exclusive.

7. BIOSTAT 3 Example: Investigation of the relationship of family history and the age at onset of bipolar disorder. Study included 318 subjects who had bipolar disorder resulting in the following data:

8. BIOSTAT 3 Calculate the following probabilities –P(E) = 141/318 =.4434 –P(A) = 63/318 =.1981 –P(A/E) = 28/141 =.1986

9. BIOSTAT 3 –P(E*A) = 28/318 =.0881 or P(E*A) = P(E) * P(A/E) = (.4434)*(.1986) =.0881 –P(E/A) = 28/63 =.4444 or P(E/A) = P(E*A)/P(A) = (.0881)*(.1981) =.4447 –Why are these different??? –NOTE: Keep in mind that all of these probabilities are based on the fact that the subject does in fact have (or will have) bipolar disorder. HW PROBLEM: 3.4.1

10. Breaking News: New test for early detection of cancer has been developed. Let C = event that patient has cancer C c = event that patient does not have cancer + = event that the test indicates a patient has cancer - = event that the test indicates that patient does not have cancer Clinical trials indicate that the test is accurate 95% of the time in detecting cancer for those patients who actually have cancer: P(+/C) =.95 but unfortunately will give a “+” 8% of the time for those patients who are known not to have cancer: P(+/ C c ) =.08 -It has also been estimated that approximately 10% of the population have cancer and don’t know it yet: P(C) =.10 -You take the test and receive a “+” test results. Should you be worried? P(C/+) = ?????

11. BIOSTAT 3 What we know. P(+/C) =.95 P(+/ C c ) =.08 P(C) =.10 From these probabilities we can find P(-/C) =.05 P(-/ C c ) =.92 P(C c ) =.90 P(C/+) = ???

12. BIOSTAT 3 The probabilities P(C) and P(C C ) are called prior probabilities because they are determined prior to the decision about taking the cancer test. The conditional probability P(C | “test results”) is called a posterior probability (or revised probability), because the prior probability is revised after the results of the cancer test is known.

13. BIOSTAT 3 P(+/C) =.95 P(+/ C c ) =.08 P(C) =.10 P(-/C) =.05 P(-/ C c ) =.92 P(C c ) =.90 Calculate P(C*+) = P(C)*P(+/C) = (.1)*(.95) =.095 P(C*-) = P(C)*P(-/C) = (.1)*(.05) =.005 P(C c *+) = P(C c )*P(+/C c ) = (.9)*(.08) =.072 P(C c *-) = P(C c )*P(-/C c ) = (.9)*(.92) =.828

14. BIOSTAT 3 P(C/+) = (.095)/(.167) =.5689 P(C/-) = (.005)/(.833) =.0060 P(C c /+) = (.072)/(.167) =.4311 P(C c /-) = (.828)/(.833) =.9940

15. BIOSTAT 3 HW Problems: –Work Example 3.5.1 in text to verify you can get the correct answer. –Work Exercise 3.5.1 Note: Sensitivity = P(+/C) Specificity = P(-/C c ) Predictive Value Positive = P(C/+) Predictive Value Negative = P(C c /-)