biostat 3

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?

BIOSTAT 3• Given k possible (mutually exclusive and

collectively exhaustive) outcomes to an experiment {O1, O2, …, Ok}, 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(Oi) ≤ 1 for each i, and

(2) The sum of the probabilities of all the outcomes equals 1

• i.e. P(O1) + P(O2) + … + P(Ok) = 1

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.

BIOSTAT 3

• Conditional Probability

• Joint Probability (Multiplication rule)

• Additional rule

P( A and B) = P(A)*P(B/A) = P(B)*P(A/B) both are true

P(A or B) = P(A + B) = P(A) + P(B) – P(A and B)

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)

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.

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:

Family History of Mood Disorders Early = 18(E) Later > 18(L) Total

Negative (A) 28 35 63Bipolar disorder (B) 19 38 57Unipolar (C) 41 44 85Unipolar and bipolar (D) 53 60 113Total 141 177 318

Table 3.4.1 Frequency of Family History of Mood Disorder by Age Group Among Bipolar Subjects

BIOSTAT 3

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

Family History of Mood Disorders Early = 18(E) Later > 18(L) Total


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: 3.4.1Family History of Mood Disorders Early = 18(E) Later > 18(L) Total


Breaking News: New test for early detection of cancer has been developed.

LetC = event that patient has cancerCc = 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) = .95but unfortunately will give a “+” 8% of the time for those patients who are known not to have cancer: P(+/ Cc ) = .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/+) = ?????

BIOSTAT 3

• What we know.P(+/C) = .95 P(+/ Cc ) = .08 P(C) = .10From these probabilities we can findP(-/C) = .05 P(-/ Cc ) = .92 P(Cc) = .90

• P(C/+) = ???

Have Cancer: C Do Not Have Cander: Cc

+

-

True State of Nature

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BIOSTAT 3

• The probabilities P(C) and P(CC) 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.

BIOSTAT 3• P(+/C) = .95 P(+/ Cc ) = .08 P(C) = .10

• P(-/C) = .05 P(-/ Cc ) = .92 P(Cc) = .90

• Calculate

• P(C*+) = P(C)*P(+/C) = (.1)*(.95) = .095

• P(C*-) = P(C)*P(-/C) = (.1)*(.05) = .005

• P(Cc*+) = P(Cc)*P(+/Cc) = (.9)*(.08) = .072

• P(Cc*-) = P(Cc)*P(-/Cc) = (.9)*(.92) = .828


+

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BIOSTAT 3

• P(C/+) = (.095)/(.167) = .5689

• P(C/-) = (.005)/(.833) = .0060

• P(Cc/+) = (.072)/(.167) = .4311

• P(Cc/-) = (.828)/(.833) = .9940


+ 0.095 0.072 0.167

- 0.005 0.828 0.833

0.1 0.9 1.000


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BIOSTAT 3

• HW Problems:– Work Example 3.5.1 in text to verify you can

get the correct answer.– Work Exercise 3.5.1Note:Sensitivity = P(+/C)Specificity = P(-/Cc)Predictive Value Positive = P(C/+)Predictive Value Negative = P(Cc/-)

biostat 3

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