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
Jalal Karimi, Epidemiologist, PhD,
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
Jalal Karimi, Epidemiologist, PhD,
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
Jalal Karimi, Epidemiologist, PhD,
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
Jalal Karimi, Epidemiologist, PhD,
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
Jalal Karimi, Epidemiologist, PhD,
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
Jalal Karimi, Epidemiologist, PhD,
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
Jalal Karimi, Epidemiologist, PhD,
Community Medicine Department
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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
Jalal Karimi, Epidemiologist, PhD,
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
Jalal Karimi, Epidemiologist, PhD,
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
Jalal Karimi, Epidemiologist, PhD,
Community Medicine Department
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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
Jalal Karimi, Epidemiologist, PhD,
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
Jalal Karimi, Epidemiologist, PhD,
Community Medicine Department
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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-)
Jalal Karimi, Epidemiologist, PhD,
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 =
Jalal Karimi, Epidemiologist, PhD,
Community Medicine Department
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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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