The document discusses key concepts in probability and epidemiology, including: 1) Definitions of probability, sensitivity, specificity, positive predictive value, and negative predictive value in diagnostic testing. 2) Methods for calculating probabilities of events using formulas like multiplication rule and addition rule. 3) Examples of how to determine the probability of outcomes from throwing dice, coins, and other chance experiments. 4) Explanations of dependent vs independent events and the use of Bayes' theorem. 5) Illustrations of how test characteristics impact positive and negative predictive values.

1. B Y : J A L A L K A R I M I , M S C , P H D , S E C O N D S E S S I O N
Jalal Karimi, Epidemiologist, PhD,
Community Medicine Department
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2. • People use the term probability many times each
day.
• For example, physician says that a patient has a 50-
50 chance of surviving a certain operation.
• Another physician may say that she is 95% certain
that a patient has a particular disease
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Community Medicine Department
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3. • If an event can occur in N mutually exclusive and
equally likely ways, and if m of these possess a trait,
E, the probability of the occurrence of E is read as
• P(E) = m/N
• Example
• We toss a die, what is the probability of 4 coming
up?
Since there are 6 mutually exclusive and equally likely
outcomes out of witch 4 is only one, the probability of
4 coming up is 1/6
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Community Medicine Department
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4. • Suppose we toss 2 coins. We have following
outcomes:
• HH, TH, HT,TT (H=Head, T=tail)
• Suppose we want to know the probability of HH.
• HH being one of the four likely outcomes, the
probability of obtaining HH is 1/4
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Community Medicine Department
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5. • Suppose we throw 2 dice and we want probability of
total of 7 points. A total 7 can come in 6.
• (1-6, 2-5, 3-4, 4,3- 5-2, or 6-1). So the numerator will be 6.
• Since we have 6 sides for each die, the total number of
‘equally likely’ , ‘mutually exclusive’ outcomes is 6×6=36.
• So the chance of getting a total of 7 when we throw 2
dice is 6/36 (or 1/6)
• Suppose there is a box containing 12 red beads and 8
green beads. You shuffle it and and pick one bead with
your eyes closed. The chance of your getting a red
bead is 12/ 20
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Community Medicine Department
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6. • The estimate of probability of a specified outcome
based on a series of independent trials is given by:
• Probability =
𝑻𝒉𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒊𝒎𝒆𝒔 𝒕𝒉𝒆 𝒐𝒖𝒕𝒄𝒐𝒎𝒆 𝒐𝒄𝒄𝒖𝒓𝒓𝒆𝒅
𝑻𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒓𝒊𝒂𝒍𝒔
• Statistical probability or a posteriori probability i.e.,
after the event
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Community Medicine Department
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7. LAWS OF PROBABILITY FOR
INDEPENDENT EVENT
• Addition law
• Example
• We throw a die, what is probability of getting 2 or 4
or 6? 1/6+1/6+1/6= 3/6 = 1/2
• Multiplication law (simultaneous occurrence)
• Example
• In tossing 2 coin, what is probability of heads in both
coins?
• 1/2×1/2= 1/4
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Community Medicine Department
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8. P(A)
P(B)
P(B|A)
A and B are independent
P(B|A) = P(B)Jalal Karimi, Epidemiologist, PhD,
Community Medicine Department
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9. P(A)
A and B are mutually exclusive
The occurrence of one event precludes the
occurrence of the other
P(B)
P(A OR B) = P(A U B) = P(A) + P(B)
Addition
Rule
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10. Two events are not mutually exclusive (male
gender and blood type O).
P(M OR O) = P(M)+P(O) – P(M∩O)
= 0.50 + 0.40 – 0.20
= 0.70
Blood Group M F Total
O
A
B
AB
20
17
8
5
20
18
7
5
40
35
15
10
Total 50 50 100
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Community Medicine Department
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11. EXAMPLE:
The joint probability of being ill and eat salad
P(Ill) = 110/200 = 0.55
P(IllEat B) = 90/120 = 0.75
Then the two events are dependent
P(Ill∩Eat B) = P(Eat B)P(IllEat B)
= (120/200)(90/120) = 0.45
P(A and B) = P(A) P(B|A)
Multiplication rule,
Dependence
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Community Medicine Department
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12. CONDITIONAL PROBABILITY
• Example: bacteriuria & Pyelonephritis
Suppose it is known that 6 percent of pregnant
women attending a prenatal clinic at a large urban
hospital have bacteriuria.
Consider the two event: A, a pregnant women has
bacteriuria, and Ā,are mutually exclusive and
complementary:
P(A)= 0.06, P(Ā)= 1-0.06= 0.94
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13. BAYES’ THEOREM‫بیز‬ ‫قضیه‬
‫تشخیصی‬ ‫کارآیی‬ ‫در‬ ‫آن‬ ‫کاربرد‬
‫اوری‬ ‫باکتری‬ ‫نداشتن‬ ‫احتمال‬P(Ā)‫و‬‫اوری‬ ‫باکتری‬ ‫احتمال‬P(A)
‫نداشتن‬ ‫احتمال‬‫پیلونفریت‬P(B)‫پیلونفریت‬ ‫احتمال‬ ‫و‬P(B)
Suppose 30% of bacteriouric and 1% of non- bacteriouric pregnant
women proceed the disease:
• P(A and B)= P(BI A)/P(A)=(0.30)(0.06)=0.018
• P(Ā and B)=P(BIĀ)/P(Ā)= (0.01)(0.94)=0.0094
BIA 0.30 ‫ابتال‬‫باشد‬ ‫اوریک‬ ‫باکتری‬ ‫که‬ ‫ی‬ ‫شرط‬ ‫به‬ ‫پیلونفریت‬ ‫به‬
BIĀ 0.01 ‫ابتال‬‫نباشد‬ ‫اوریک‬ ‫باکتری‬ ‫ی‬ ‫شرط‬ ‫به‬ ‫پیلونفریت‬ ‫به‬
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Community Medicine Department
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14. •‫باشد‬ ‫داشته‬ ‫پیلونفریت‬ ‫هم‬ ‫و‬ ‫اوری‬ ‫باکتری‬ ‫هم‬ ‫باردار‬ ‫زن‬ ‫اینکه‬ ‫شانس‬:
• P(A and B)= P(BI A)/P(A)=(0.30)(0.06)=0.018
•‫اوری‬ ‫باکتری‬ ‫هم‬ ‫باردار‬ ‫زن‬ ‫اینکه‬ ‫شانس‬‫پیلونفریت‬ ‫و‬ ‫نداشته‬‫باشد‬ ‫داشته‬:
• P(Ā and B)=PBIĀ)/P(Ā)= (0.01)(0.94)=0.0094
•‫پیلونفریت‬ ‫شانس‬:
P(‫باشد‬ ‫داشته‬ ‫=)پیلونفریت‬P(B)= P(A and B)+ P(Ā and B)
=(0.018)(0.0094)=0.0274
‫باشد‬ ‫داشته‬ ‫پیلونفریت‬ ‫باردار‬ ‫زن‬ ‫که‬ ‫شرطی‬ ‫به‬ ‫اوری‬ ‫باکتری‬ ‫شانس‬:
P(AIB)=P(A and B)/P(B)= 0.0180/0.0274= 0.6569
‫پس‬:‫دارد‬ ‫باردارپیلونفریت‬ ‫زن‬ ‫اگر‬65.7%‫باشد‬ ‫داشته‬ ‫اوری‬ ‫باکتری‬ ‫دارد‬ ‫احتمال‬
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15. • Sensitivity: The ability to detect people who do
have disease (T+ | D+)
• Specificity: The ability to detect people who do not
have disease (T- | D-)
• Positive Predictive Value: The likelihood that a
person with a positive test result actually has disease
(T+ | D+)
• Negative Predictive Value: The likelihood that a
person with a negative test result truly does not
have disease (T-|D-)
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Community Medicine Department
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16. Disease
Positive
Disease
Negative
Test Positive True Positive False Positive
Test Negative False Negative True Negative
True Positive
True Positive+ False Negative
Sensitivity =
True Negative
True Negative+ False Positive
Specificity =
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17. ‫تست‬ ‫بیماری‬
‫دارد‬ ‫ندارد‬
‫مثبت‬ 9 100 109
‫منفی‬ 1 9890 9891
10 9990 10000
Test Disease Total
+ _
+ a b a + b
- c d c + d
Total a+c b+d a+b+c+d
Sensitivity= 9/10= 0.9= 90% Specificity= 9890/9990= 99=99%Jalal Karimi, Epidemiologist, PhD,
Community Medicine Department
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18. ‫تست‬ ‫بیماری‬
‫دارد‬ ‫ندارد‬
‫مثبت‬ 900 90 990
‫منفی‬ 100 8910 9010
1000 9000 10000
Sensitivity = 900/1000=0.9 = 90%
Specificity= 8910/9000 =0.99= 99%
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Community Medicine Department
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19. POSITIVE PREDICTIVE VALUE
Disease Positive
Disease
Negative
Test Positive True Positive False Positive
Test Negative False Negative True Negative
True Positive
True Positive+ False Positive
=‫مثبت‬ ‫اخباری‬ ‫ارزش‬
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Community Medicine Department
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20. NEGATIVE PREDICTIVE VALUE
Disease Positive Disease Negative
Test Positive True Positive False Positive
Test Negative False Negative True Negative
True Negative
True Negative+ False Negative
=‫منفی‬ ‫اخباری‬ ‫ارزش‬
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Community Medicine Department
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21. SENSETVIITY=90
SPECIFICITY = 99
‫تست‬ ‫بیماری‬
‫کل‬‫دارد‬ ‫ندارد‬
‫مثبت‬ 9 100 109
‫منفی‬ 1 9890 9891
10 9990 10000
‫تست‬ ‫بیماری‬
‫کل‬‫دارد‬ ‫ندارد‬
‫مثبت‬ 900 90 990
‫منفی‬ 100 8910 9010
1000 9000 10000
=‫ارزش‬
‫مثبت‬ ‫اخباری‬
=‫ارزش‬
‫منفی‬ ‫اخباری‬
‫واقعی‬ ‫مثبت‬
‫کاذب‬ ‫منفی‬+‫واقعی‬ ‫مثبت‬
‫واقعی‬ ‫منفی‬
‫کاذب‬ ‫مثبت‬+‫واقعی‬ ‫منفی‬
(8)%0.08=9/109=
(99.9)%0.99=9890/9891=
PPV (90)%0.9=900/990=
NPV (99)%0.99=8910/9010=
‫نکته‬:
‫بیماری‬ ‫شیوع‬ ‫هرچه‬
‫ارزش‬ ،‫باشد‬ ‫بیشتر‬
‫اخباری‬ ‫ارزش‬ ‫و‬ ‫بیشتر‬ ‫تست‬ ‫مثبت‬ ‫اخباری‬
‫منفی‬
‫است‬ ‫کمتر‬ ‫تست‬.Jalal Karimi, Epidemiologist, PhD,
Community Medicine Department
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22. •Sensitivity
•‫هستند‬ ‫بیماری‬ ‫دارای‬ ‫که‬ ‫است‬ ‫افرادی‬ ‫تشخیص‬ ‫در‬ ‫آزمایش‬ ‫یک‬ ‫توانایی‬
•Specificity
•‫ندارند‬ ‫را‬ ‫بیماری‬ ‫که‬ ‫است‬ ‫افرادی‬ ‫تشخیص‬ ‫در‬ ‫آزمایش‬ ‫یک‬ ‫توانایی‬
•Positive predictive value
•‫باشند‬ ‫مثبت‬ ً‫ا‬‫واقع‬ ‫که‬ ‫است‬ ‫مثبت‬ ‫نتایج‬ ‫از‬ ‫نسبتی‬
•Negative predictive value
•‫از‬ ‫نسبتی‬‫ن‬‫باشند‬ ‫منفی‬ ً‫ا‬‫واقع‬ ‫که‬ ‫است‬ ‫منفی‬ ‫تایج‬
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Community Medicine Department
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23. End of second session
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Community Medicine Department
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