This document discusses various laws of probability including addition law, multiplication law, and binomial law. It provides examples of how to calculate probabilities using these laws, such as calculating the probability of mutually exclusive events using addition law. It also discusses key probability concepts like marginal probability, joint probability, conditional probability, sensitivity, specificity, predictive values, and likelihood ratios.

1. LAWS OF
PROBABILITY
Presenter
Dr. Brijesh Kumar
JR
Department of community Medicine, PGIMS, Rohtak

2. Contents
Definition
Types of probability
Bayes’ theorem
Binomial Probability Law
Laws of probability
References

3. Probability
Relative frequency or probable chances of
occurrence with which an event is expected to
occur on an average.
Odds with which an event is expected to occur
in a long run.
OR

4. Usually expressed as symbol ‘p’
Probability
‘p’ ranges from 0 to 1
P=0 means ‘ no chance of an event happening’
P=1 means ‘100% chances of an event
happening’

5. p = no. of events occurring
total no. of trials
p + q = 1
Probability
If probability of an event happening is ‘p’ and
probability of not happening is ‘q’
If an experiment is repeated ‘n’ times and an
event A is observed ‘f’ times
p(A) = f/n

6. Laws of Probability
Addition law of Probability
Multiplication law of Probability
Binomial law of Probability

7. Addition law of Probability
If one event excludes the possibility of
occurrence of the other specified event or
events, the events are called mutually
exclusive.
e.g. Getting head excludes the possibility of
getting tail in coin flip
e.g. Birth of a male child excludes the
possibility of birth a female

8. Addition law of Probability
These mutually exclusive events follow the
addition law of probability.
If the no. of mutually exclusive events are n
and p1 is the individual probability
Total prob. P = p1+ p2 + ………….+ pn = 1
P(A or B) = P(A) + P(B)

9. Addition law of Probability
P(2) = 1/6
P(5) = 1/6
P(2 or 5) = 1/6 + 1/6 = 1/3
P(aces of any colour) =
1/52 +1/52 + 1/52 + 1/52
= 1/13

10. Multiplication law of Probability
It is applied when two or more events are
occurring together but they are independent of
each other
If the no. of events are n and p1 is the
individual probability
Total prob. P = p1 * p2 * ………….* pn
P(A and B) = P(A) * P(B)

11. P(2) = 1/6
P(5) = 1/6
P(2 and 5) = 1/6 * 1/6 = 1/36
Multiplication law of Probability

12. Multiplication law of Probability
If we assume twin pregnancy will occur once in
80 preg. And child with Rh-ve blood group will be
born once in 10 births.
prob. of male child,p1 = ½
prob. of child being Rh+ve, p2 = 9/10
prob. of single birth, p3 = 79/80
prob. of male, Rh+ve and single birth =
½*9/10*79/80 = 711/1600

13. Binomial law of Probability
It is used when events are occurring one after the
other.
e.g.
M
M
F
F
M
F
M Mand
F
F
F
and
and
and
M
M
F
1/2*1/2=1/4
1/2*1/2=1/4
1/2*1/2=1/4
1/2*1/2=1/4

14. M
M
F
F
M
F
M Mand
F
F
F
and
and
and
M
M
F
1/2*1/2=1/4
1/2*1/2=1/4
1/2*1/2=1/4
1/2*1/2=1/4
Binomial law of Probability
Chances of getting two males = ¼ = 25%

15. Binomial law of Probability
M
M
F
F
M
F
M Mand
F
F
F
and
and
and
M
M
F
1/2*1/2=1/4
1/2*1/2=1/4
1/2*1/2=1/4
1/2*1/2=1/4
Chances of getting two females = ¼ = 25%

16. Binomial law of Probability
M
M
F
F
M
F
M Mand
F
F
F
and
and
and
M
M
F
1/2*1/2=1/4
1/2*1/2=1/4
1/2*1/2=1/4
1/2*1/2=1/4
Chances of getting one of either sex = 1/4 + 1/4
= ½ = 50%

17. Binomial law of Probability
M
M
F
F
M
F
M
F
F
M
M
M
F
F
e.g. MMM
MFM
MMF
MFF
FMM
FMF
FFM
FFF
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8

18. Binomial law of Probability
M
M
F
F
M
F
M
F
F
M
M
M
F
F
MMM
MFM
MMF
MFF
FMM
FMF
FFM
FFF
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
Chances of getting 3 males = 1/8 = 12.5%

19. Binomial law of Probability
M
M
F
F
M
F
M
F
F
M
M
M
F
F
MMM
MFM
MMF
MFF
FMM
FMF
FFM
FFF
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
Chances of getting 3 Females = 1/8 = 12.5%

20. Binomial law of Probability
M
M
F
F
M
F
M
F
F
M
M
M
F
F
MMM
MFM
MMF
MFF
FMM
FMF
FFM
FFF
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
Chances of getting 2 males and 1 female =
1/8+1/8+1/8 = 37.5%

21. Binomial law of Probability
M
M
F
F
M
F
M
F
F
M
M
M
F
F
MMM
MFM
MMF
MFF
FMM
FMF
FFM
FFF
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
1/2*1/2*1/2=1/8
Chances of getting 1 male and 2 females =
1/8+1/8+1/8 = 37.5%

22. Binomial law of Probability
Binomial law of probability distribution is formed
by terms of the expansion of binomial expression
(p +q)n
n= number of events
p= probability of success
q= probability of failure

23. Binomial law of Probability
when n = 2
( p + q )2 = p2 + q2 + 2pq
when n = 3
( p + q )3 = p3 + q3 + 3pq2 + 3p2q
when n = 4
( p + q )4 = p4 + 4p3q+ 6p2q2 + 4pq3 + q4

24. If p = chance of getting male child
q = chance of getting female child
( p + q )2 = p2 + q2 + 2pq
= 0.52 + 0.52 + 2*0.5*0.5
= 0.25 + 0.25 + 0.50
Binomial law of Probability
( p + q )3= p3 + q3 + 3pq2 + 3p2q
= 0.53 + 0.53 + 3*0.5*0.52 + 3*0.52*0.5
= 0.125 + 0.125 + 0.375 + 0.375

25. Binomial law of Probability
e.g.
In 4 ICDS projects, 20% of children under 6 yrs of
age were found to be severely malnourished. Only
4 children were selected at random from the 4
projects.
What is the prob. Of 4, 3, 2, 1, 0 being severely
malnourished?

26. Binomial law of Probability
p = 0.2 q = 0.8 n = 4
( p + q )4 = p4 + 4p3q+ 6p2q2 + 4pq3 + q4
= 0.24 + 4*0.23*0.8+ 6*0.22*0.82 +
4*0.2*0.83 + 0.84
= 0.0016 + 0.0256 + 0.1536 + 0.4096 +
0.4096
Prob. of all 4 children being malnourished = 0.16%
Prob. of 3 children being malnourished = 2.56%
Prob. of 2 children being malnourished = 15.36%
Prob. of 1 children being malnourished = 40.96%
Prob. of no children being malnourished = 40.96%

27. Types of Probability
Marginal Probability
Joint Probability
Conditional Probability

28. Marginal Probability
Marginal probability is the probability of a single
event without consideration of any other event.
Written as P( A and B) or P( A B )
Joint Probability
The probability of the intersection of two events is
called their joint probability.

29. Conditional Probability
Conditional probability is the probability that an
event will occur given that another has already
occurred.
If A and B are two events,
then prob. Of A given that B has already occurred
will be :
P(A|B) = P(A and B) / P(B)
P(B|A) = P(A and B) / P(A)

30. Types of Probability
Heart
disease
cancer diabetes total
Family
History
88 44 38 170
No family
history
92 76 62 230
total 180 120 100 400
e.g.

31. Heart
disease
cancer diabetes total
Family
History
88 44 38 170
No family
history
92 76 62 230
total 180 120 100 400
Types of Probability
P ( cancer ) = 120/400 = 0.30
Marginal Probability

32. Types of Probability
Heart
disease
cancer diabetes total
Family
History
88 44 38 170
No family
history
92 76 62 230
total 180 120 100 400
P ( Cancer and No history ) = 76/400 = 0.19
Joint Probability

33. Types of Probability
Heart
disease
cancer diabetes total
Family
History
88 44 38 170
No family
history
92 76 62 230
total 180 120 100 400
P ( Diabetes | History ) = P(Dia. and Hist.)/ P(Hist.)
= 38/170 = 0.2235
Conditional Probability

34. Conditional Probability
Test Result Disease
present
(D+)
Disease
absent
(D-)
Total
Positive ( T+) a (TP) b (FP) a+b
Negative (T-) c (FN) d (TN) c+d
Total a+c b+d n
Sensitivity = prob. of positive test given the
presence of disease = P ( T+ | D+ )
= P ( T+ and D+ )/ P( D+)
= a/ (a+c)

35. Conditional Probability
Test Result Disease
present
(D+)
Disease
absent
(D-)
Total
Positive ( T+) a (TP) b (FP) a+b
Negative (T-) c (FN) d (TN) c+d
Total a+c b+d n
Specificity = prob. of negative test given the
absence of disease = P ( T- | D- )
= P ( T- and D- )/ P( D-)
= d/ (b+d)

36. Conditional Probability
Test Result Disease
present
(D+)
Disease
absent
(D-)
Total
Positive ( T+) a (TP) b (FP) a+b
Negative (T-) c (FN) d (TN) c+d
Total a+c b+d n
Positive predictive value = prob. of presence of
disease given the positive test = P ( D+ | T+ )
= P ( T+ and D+ )/ P( T+)
= a/ (a+b)

37. Conditional Probability
Test Result Disease
present
(D+)
Disease
absent
(D-)
Total
Positive ( T+) a (TP) b (FP) a+b
Negative (T-) c (FN) d (TN) c+d
Total a+c b+d n
Negative predictive value = prob. of absence of
disease given the negative test = P ( D- | T- )
= P ( T- and D- )/ P( T-)
= d/ (c+d)

38. Likelihood Ratios
There are two types of Likelihood Ratios one for
positive test (LR+) and another negative test
(LR-).
LR is the ratio of probability of a particular test
result for a individual with disease of interest to
probability of a particular test result for a individual
without disease of interest.

39. New
Diagnostic
test
Western
Blot
pos.(D+)
Western
Blot neg.
(D-)
Total
HIV + ( T+) 85 20 105
HIV - (T-) 15 180 195
Total 100 200 300
Likelihood Ratios
e.g.
P(T+|D+)=P(T+ and D+)/P(D+) = 85/100
P(T+|D-)=P(T+ and D-)/P(D-) = 20/200

40. P(T+|D+)
LR+ = -------------
P(T+|D-)
P(T-|D+)
LR- = -------------
P(T-|D-)
D+ denotes that the individual is having a disease
by Gold Standard.
D- denotes that the individual is not having a
disease by Gold Standard.
T+ denotes that the individual is having a disease
by New diagnostic kit.
T- denotes that the individual is not having a
disease by New diagnostic kit.
Likelihood Ratios

41. Likelihood Ratios
LR+ = P(T+ | D+ ) / P(T + | D-)
LR+ = (85/100) / (20/200)
P(T+|D+) = 8.5 times P(T+|D-)
LR+ = 8.5

42. Bayes’ Theorem
The concept of conditional probability can be revised
based on new information and to determine the
probability that a particular effect was due to specific
causes. The procedures for revising these
probabilities is known as Bayes’ theorem.

43. P (A|B) = P(A and B ) / P(B)
P(A and B) = P(A|B) * P(B)
Bayes’ Theorem
P (B|A) = P(A and B ) / P(A)
P(A and B) = P(B|A) * P(A)
P (B|A) * P(A) = P(A|B) * P(B)
P(B|A) = P(A|B) * P(B) / P(A)

44. Bayes’ Theorem
e.g.
Statistics show that 60% of all men have some form
of prostate cancer. PSA test will show positive 80%of
the time for inflicted men and 5% for healthy men.
What is the probability that a man who tested
positive does have prostate cancer?????

45. Bayes’ Theorem
P(C) = 0.60 P(C’) = 0.40
P(Test positive|Cancer) = P(T|C) = 0.80
P(Test positive|No Cancer)= P(T|C’) = 0.05
Calculate P(C|T) ????

46. Bayes’ Theorem
P(C|T) = P(T|C) * P(C) / P(T)
P(T) = P(C and T) + P(C’ and T)
= P(C) * P(T|C) + P(C’) * P(T|C’)
= 0.6*0.8 + 0.4*0.05
= o.48 + 0.02 = 0.50
P(C|T) = 0.8 * 0.6 / 0.5 = 0.96
Chance of prostate cancer if tested positive is 96%

47. Binomial Probability Law
Probability of exactly ‘r’ successes in ‘n’
independent trials is given by
P(r) = n! / r!(n-r)! * pr q(n-r)
where p = prob. of success in each trial
q = prob. of failure in each trial

48. Binomial Probability Law
Trials are identical i.e. all the repetitions are
performed under identical conditions.
Trials are independent i.e. outcome of one trial
does not affect the outcome of another trial.
Each trial results in two mutually exclusive
outcomes , success (p) and failure (q).
Prob. of ‘p’ and ‘q’ remains constant for each trial.

49. Binomial Probability Law
e.g.
In a district survey,
%age of NVD = 68% i.e. P(V)
%age of del. conducted in health inst.=93% i.e P(H)
If a sample of 10 del. cases are randomly selected,
find the prob. that 6 cases were having NVD.
And find the prob. of 6 cases delivered in health
inst.

50. Binomial Probability Law
p= 0.68 P(V)
q= 0.32
P(r) = n! / r!(n-r)! * pr q(n-r)
P(6) = 10! / 6! (10-6)! * 0.686 0.324
= 0.2177

51. p= 0.93 P(H)
q= 0.07
P(r) = n! / r!(n-r)! * pr q(n-r)
P(6) = 10! / 6! (10-6)! * 0.936 0.074
= 0.0033
Binomial Probability Law

52. Further readings….
Mahajan’s Methods in Biostatistics
Handbook of statistics : Dr. Basnnar
Basic and Clinical Biostatistics : Dawson
and Trapp
S.P.Gupta’s textbook of statistics

53. Thank
You