Probability and Probability
Distribution
Dr Manoj Kumar Bhambu
GCCBA-42, Chandigarh
M- +91-988-823-7733
mkbhambu@hotmail.com
Objectives of Learning this Unit
 Probability
and
Probability
Distribution:
Definitions- Probability Rules –Application of
Probability
Rules-Conditional
ProbabilityBayes theorem
 Random Variable and Probability Distributions;
BinomialDistribution- Poisson Distribution and
Normal Distribution.
Probability: Meaning
 A probability is a quantitative measure of uncertainty –
a number that conveys the strength of our belief in the
occurrence of an uncertain event.
 Probability is a measure of uncertainty of events in a
random experiment.
Types of Experiment
1. Deterministic Experiment: Experiments that
have fixed outcome or result no matter any
number of times they are repeated
2. Random Experiment: if an experiment, when
repeated under identical conditions, do not
produce the same outcome every time but the
outcome in a trial is one of the several possible
outcomes then such an experiment is known as
random or probabilistic experiment.
Basic Terms in Probability
 A Random Experiment is any situation whose
outcome cannot be predicted with certainty.
Examples of an experiment include rolling a die,
flipping a coin, and choosing a card from a deck
of playing cards.
 By an outcome or simple event we mean any
result of the experiment. For example, the
experiment of rolling a die yields six outcomes,
namely, the outcomes 1,2,3,4,5, and 6.
Examples of ‘Outcomes’
 If you toss a coin you have an equal chance of
getting a head or a tail. Heads or tails are the
‘outcomes’.
 If a baby is born it has an equal chance of being a
boy or a girl. Boy or girl are the ‘outcomes’.
 If you roll a dice you have an equal chance of
getting the number 1, 2, 3, 4, 5 or 6! These scores
are the ‘outcomes’.
Basic Terms in Probability (con.)
 The sample space S of an experiment is the set of all
possible outcomes for the experiment. For example, if you
roll a die one time then the experiment is the roll of the
die. A sample space for this experiment could be S = {1, 2,
3, 4, 5, 6} where each digit represents a face of the die.
 Discrete Sample Space: A sample space whose elements are
finite or infinite but countable is called Discrete Sample
Space
 Continuous Sample Space: A Sample Space whose elements
are infinite and uncountable or assume all the values on a
real line R or on an intervals of R is called a Continuous
Sample Space.
 Event: An event is a subset of the sample space. For
example, the event of rolling an odd number with a
die consists of three simple events {1, 3, 5}.
 Null Event: An event having no sample point is called
a Null Event and is denoted by Φ.
 Simple Event: An event consisting of only one
sample point of a sample space is called a simple
event.
 Compound Events: When an event is decomposable
into a number of simple events, then it is called a
compound event.
 Exhaustive Events: It is the total number of all
possible outcomes of an experiment.
 Mutually Exclusive Events: If in an experiment the
occurrence of an event prevents or rules out the
happening of all other events in the same experiment,
then these events are said to be mutually exclusive
events.
 Equally Likely Events: The outcomes of an experiment
are equally likely to occur when the probability of
each outcome is equal.
 Collectively Exhaustive Events : The total number of
events in a population exhausts the population. So
they are known as collectively exhaustive events.
 Favourable cases: The cases which ensures the
occurrence of an event are said to be favourable
cases to the event.
Independent and Dependent Events
 When an experiment is conducted in such a way that the
occurrence of event neither effect the occurrence of
another event nor is effected by the occurrence of
another event, both the events are called independent
events.
 Events which are not independent are called dependent
events
Mathematical Vocabulary
 We can describe the probability or chance of an
event happening by saying:
 It is IMPOSSIBLE.
 It is UNLIKELY.
 It is LIKELY.
 It is CERTAIN.
Classical Definition of Probability
 If an experiment has n mutually exclusive, equally likely
and exhaustive cases, out of which m are favourable to
the happening of event A, then the probability of the
happening of A is denoted by P(A) and is defined as:
P(A) =
𝑚
𝑛
=
𝑁𝑜 𝑜𝑓 𝐶𝑎𝑠𝑒𝑠 𝑓𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑡𝑜 𝐴
𝑇𝑜𝑡𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑎𝑠𝑒𝑠
 Probability of an event which is certain to occur is 1
and the probability of an event impossible is zero.
 The probability of occurrence of any event lies
between 0 and 1, both inclusive
Example 1
 What is the probability of getting an even number in a
single throw of a dice.
 Answer: The possible cases in the throw of a dice are
six viz. 1, 2, 3, 4, 5, 6.
 Favourable cases are those which are marked with 2,
4, and 6.
 ∴ Probability of getting an even no =
𝑚
𝑛
=
3 1
=
6 2
More Examples
 What is the probability of getting tail in a throw of a
unbiased coin?
 A bag contains 6 white balls and 9 black balls. What is the
probability of drawing a black ball in a single draw?
 What is the probability when a card is drawn at random
from an ordinary pack of cards, if it is (i) a red card; (ii) a
club; (iii) one of the court cards?
 What is the probability of throwing a number greater
than 3 with an ordinary dice?
 What is the probability that a leap year selected at
random will have 53 Sundays?
 What is the probability of getting a total of more than 10
in a single throw of two dice?
Statistical or Empirical definition of
Probability
 “ if the experiment be repeated a large number of
times under essentially identical conditions, the
limiting value of the ratio of the number of times
the event A happens to the total number of trials of
the experiments as the number of trials increases
indefinitely is called the probability of happening of the
event A.”
-Von Mises
 Symbolically, Let P(A) denote the probability of occurrence
of event A, m be the number of times in which the event A
occurs in a series of n trials then
𝑚
𝑛→∞ 𝑛
P(A)= lim
provided the limit is finite and unique
Modern Definition of Probability
 Definition. A probability function P on a finite sample
space Ω assigns to each event A in Ω a number P(A) in
[0,1] such that
(i) P(Ω) = 1, and
(ii) P(A ∪ B) = P(A) + P(B) if A and B are disjoint.
The number P(A) is called the probability that A
occurs.
Greek Alphabets
Theorems on Probability
 There are two important theorems of probability, namely
1. Addition Theorem or Theorem on Total Probability
2. Multiplication Theorem or Theorem on Compound
Probability
Addition Theorem
 Statement: If n events are mutually exclusive, then the
probability of happening of any one of them is equal to the
sum of the probabilities of the happening of the separate
events. In other words, if 𝐸1 , 𝐸2, 𝐸3, …….. 𝐸𝑛 be n events and
P( 𝐸1 ), P( 𝐸2, ), P( 𝐸3, ), …….P( 𝐸𝑛 ), be their respective
probabilities, then
P(𝐸1 + 𝐸2 +𝐸3 ……..+ 𝐸𝑛 )=P(𝐸1 ) + P(𝐸2 )+𝑃(𝐸3 )+……..+P( 𝐸𝑛 )
Proof: Let 𝐸1 , 𝐸2, 𝐸3, …….. 𝐸𝑛 be mutually exclusive events with
probabilities P(𝐸1 ), P(𝐸2, ), P(𝐸3, ), …….P(𝐸𝑛 ) respectively. Let N be
the total number of trials. Let 𝑚1 , 𝑚2, 𝑚3, …….. 𝑚𝑛 be the cases
favourable to events 𝐸1 , 𝐸2, 𝐸3, …….. 𝐸𝑛 respectively, then
𝑚
𝑁
P(𝐸𝑖 ) = 𝑖 , i= 1, 2, ………, n.
Since the events 𝐸1 , 𝐸2, 𝐸3, …….. 𝐸𝑛 are mutually exclusive,
therefore, the cases favourable to the happening of any one of the
events 𝐸1 , 𝐸2, 𝐸3, …….. 𝐸𝑛 are 𝑚1 , 𝑚2, 𝑚3, …….. 𝑚𝑛 . Hence the
probability of happening of any one of the events 𝐸1 , 𝐸2, 𝐸3, ……..
𝐸𝑛 is
P(𝐸1 + 𝐸2 +𝐸3 +……..+ 𝐸𝑛 ) =
𝑚1+ 𝑚2+ …….+𝑚𝑛
𝑁
=
𝑚1
𝑁
+
𝑚2
𝑁
+ ...... +
=P(𝐸1 ) + P(𝐸2 )+𝑃(𝐸3 )+……..+P( 𝐸𝑛 )
𝑚𝑛
𝑁
Example
 A dice is rolled. What is the probability of getting a 1 or 6?
 Answer: Let 𝐸1 , 𝐸2, be the event of getting a 1 and 6
respectively when a dice is rolled.
Here P(𝐸1 )=
1
6
P(𝐸2 ) =
1
6
∴ P(𝐸1 + 𝐸2 ) = P(𝐸1 )+ P(𝐸2 ) ( By Addition Theorem)
1
6
1
6
2
6
= + = =
1
3
Example
 If the probability of horse A winning the race is
1
,
6
1
5
and the
probability of horse B winning the race is
what is the
probability that one of the horse will win the race?
 Answer: Let 𝐸1 , 𝐸2, be the event of winning the race by horse A
and horse B respectively.
Here P(𝐸1 )=
1
5
P(𝐸2 ) =
1
6
∴ P(𝐸1 + 𝐸2 ) = P(𝐸1 )+ P(𝐸2 ) ( By Addition Theorem)
=
1 1 6+5
+ =
5 6
30
11
=
30
EXAMPLES ILLUSTRATING THE APPLICATION OF THE ADDITION
THEOREM
Question: A card is drawn from a pack of 52 cards. What is the
probability of getting either a king or a queen?
Solution: There are 4kings and 4 queens in a pack of 52 cards.
The probability of drawing a king card is P (K) =
4
52
and the probability of drawing a queen card is P (Q) =
4
52
Since, both the events are mutually exclusive, the probability that the card drawn
is either a king or a queen is
P (K or Q) = P (K) + P (Q)
( By Addition Theorem)
=
4
4
+
52 52
=
8
2
=
52 13
More Example
 Question :A card is drawn at random from a pack of cards. Find
the probability that the drawn card is either a club or an ace of
diamond.
13
 Solution :Probability of drawing a club P (A) =
52
1
Probability of drawing an ace of diamond P (B) =
52
P (A or B) = P (A) + P (B)
13
1
14
7
+ = =
52 52 52 26
(By Addition Theorem)
Question: An investment consultant predicts that the odds
against the price of a certain stock will go up during the next
week are 2 : 1 and the odds in favour of the price remaining
the same are 1 : 3. What is the probability that the price of the
stock will go down during the next week.
Solution: Let A denote the event ‘stock price will go up’, and B
be the event ‘stock price will remain same’.
𝟏
𝟏
Then P (A) = , and P (B) =
𝟑
𝟒
P (stock price will either go up or remain same) = P (A U B)
𝟏 𝟏
𝟕
P (A) + P (B) = + =
𝟑 𝟒
𝟏𝟐
Now, P (stock price will go down) = 1 – P (A U B)
=1-
𝟕
𝟏𝟐
=
𝟓
𝟏𝟐
ADDITION THEOREM FOR NOT MUTUALLY EXCLUSIVE EVENTS
Two or more events are known as partially overlapping if part of one event
and part of another event occur together. Thus, when the events are not
mutually exclusive the addition theorem has to be modified.
Modified Addition Theorem states that if A and B are not mutually exclusive
events, the probability of occurrence of either A or B or both is equal to the
probability of that event A occurs, plus the probability that event B occurs
minus the probability that events common to both A and B occur
simultaneously. Symbolically,
P (A or B or Both ) = P (A) + P (B) – P (AB)
The following figure illustrates this point:
NOT MUTUALLY EXCLUSIVE EVENTS
A
AB
B
Overlapping Events
Generalization
The theorem can be extended to three or more events. If A, B and C are not mutually
exclusive events, then
P (Either A or B or C) = P (A) + P (B) + P (C) – P (AB) – P (AC) – P (BC) + P (ABC)
EXAMPLES ILLUSTRATING THE APPLICATION OF THE MODIFIED ADDITION THEOREM
Question: A card is drawn at random from a well shuffled pack of cards. What is the
probability that it is either a spade or a king ?
Solution: The Probability of drawing a spade P (A) =
The Probability of drawing a King P (B) =
4
52
The Probability of drawing a King of Spade P (AB) =
P ( A or B or Both) = P (A) + P (B) – P (AB)
=
=
16
4
=
52 13
13
4
1
+ 52 52 52
13
52
1
52
Question: A bag contains 30 balls number from 1 to
30. One ball is drawn at random. Find the probability
that the number of ball is a multiple of 5 or 6.
Solution: Probability of the ball being a multiple of 5 P
6
(A) =
30
5
Probability of the ball being a multiple of 6 P (B) =
30
Since, 30 is a multiple of 5 as well as 6, therefore the
events are not mutually exclusive.
1
P (A and B) =
(common multiple 30)
30
P (A or B) = P (A) + P (B) – P (AB)
6
5
1
1
= + - =
30
30
30
3
MULTIPLICATION THEOREM: Independent Events
The Multiplication Theorem states that if A and B are two
independent events, then the probability of the simultaneous
occurrence of A and B is equal to the product of their individual
probabilities. Symbolically,
P (AB) = P (A) X P (B)
Let 𝑚1 be the number of cases favourable to the happening of the
event A out of 𝑛1 exhaustive and equally likely cases.
𝑚
P (A) = 1
𝑛1
Let 𝑚2 be the number of cases favourable to the happening of the
event B out of 𝑛2 exhaustive and equally likely cases.
𝑚
P (B) = 2
𝑛2
Now, by the Fundamental Principle of counting , the number of
cases favourable to the happening of the event AB is 𝑚1 𝑚2 out of
𝑛1 𝑛2
𝑚 𝑚
P (AB) = 1 2
𝑛1 𝑛2
P (AB) = P (A).P (B)
Hence, the theorem is proved.
Generalisation
The theorem can be extended to three or more independent events. If A, B and C are three
independent events, then
P (ABC) = P (A) x P (B) x P (C)
EXAMPLES ILLUSTRATING THE APPLICATION OF THE MULTIPLE
THEOREM
Question: A coin is tossed three times. What is the probability of getting all the three heads?
1
Solution: Probability of head in the first toss P (A) =
2
1
Probability of head on the second toss P(B) =
2
1
Probability of head on the third toss P(C) =
2
P (ABC) = P (A) x P (B) x P(C)
1 1 1
= x x
2 2 2
1
=
8
Question: From a pack of 52 cards, two cards are drawn
at random one after the another with replacement.
What is the probability that both cards are kings?
Solution: The probability of drawing a King P (A) =
The probability of drawing
4
replacement P (B) =
52
P (AB) = P (A) x P (B)
=
4
52
x
4
52
=
1
169
again
a
4
52
king
after
Probability of happening of atleast one event in case of n
independent events
P (happening of at least one of the events) = 1 – P (happening of none of the
events)
Question: A problem in statistics is given to three students. A, B and C whose chances
od solving it are ½, 1/3 and ¼. What is the probability that the problems will be solved?
1
Solution: Probability that A will solve the problem = P (A) =
2
1
Probability that B will solve the problem = P (B) =
3
1
Probability that C will solve the problem = P (C) =
4
1 1
Probability that A will not solve the problem = P (𝐴 ) = 1 - =
2 2
1 2
Probability that B will not solve the problem = P (𝐵 ) = 1 - =
3 3
1 3
Probability that C will not solve the problem = P (𝐶 ) = 1 - =
4 4
P (that none will solve the problem) = P (𝐴 ).P (𝐵 ).P (𝐶 )
1 2 3 1
= x x =
2 3 4 4
P (that problem will be solved) = 1 – P (that none will solve)
1 3
=1- =
4 4
Question: A candidate (Mr. X) is interviewed for 3 posts. For the first post, there are 3
candidates, for the second post, there are 4 and for the third there are 2.
What are
the chances of Mr. X being selected ?
1
Solution : Probability of selection for 1st post = P (A) =
3
1
Probability of selection for 2nd post = P (B) =
4
1
rd
Probability of selection for 3 post = P (C) =
2
1 2
st
Probability of not getting selected for 1 post = P (𝐴) = 1 - =
3 3
1
3
Probability of not getting selected for 2nd post = P (𝐵) = 1 =
4
4
1 1
Probability of not getting selected for 3rd post = P (𝐶) = 1 - =
2 2
P (𝐴𝐵 𝐶) = P (𝐴 ).P (𝐵 ).P (𝐶 )
2 3 1 1
= x x =
3 4 2 4
Probability of selection for at least 1 post
= 1 – P (not selected at all)
1 3
=1- =
4 4
CONDITIONAL PROBABILITY
Dependent events are those in which the occurrence of one event affects the probability of
other events. The probability of the event B given that A has occurred is called conditional
probability of B. It is denoted by P (b/A). Similarly, the conditional probability of A given
that B has occurred is denoted by P (A/B).
If A and B are two dependent events, then the conditional probability of B given A is
defined and given by:
𝑃 (𝐴𝐵)
P (B/A) =
𝑃 (𝐴)
provided P (A) > 0
Similarly, the conditional probability of A given B is defined and given by:
P (A/B) =
𝑃 (𝐴𝐵)
𝑃 (𝐵)
provided P (B) > 0
MULTIPLICATION THEOREM IN CASE OF DPENDENT VARIABLES
When the events are not independent, i.e., they are dependent events, then the
multiplication theorem has to be modified. The Modified Multiplication Theorem states that
if A and B are two dependent events, then the probability of their simultaneous occurrence is
equal to the probability of one event multiplied by the conditional probability of the other.
P (AB) = P (A) . P (B/A)
OR
P (AB) = P (B) . P (A/B)
Where, P (B/A) = Conditional Probability of B given A.
P (A/B) = Conditional Probability of A given B.
EXAMPLES ILLUSTRATING THE APPLICATION OF THE MODIFIED
MULTIPLICATION THEOREM
Question: A bag contains 10 white and 5 black balls. Two
balls are drawn at random one after the other without
replacement. Find the probability that both balls drawn are
black.
Solution:
The probability of drawing a black ball in the first attempt is:
5
5
P (A) =
=
10+5
15
The probability of drawing the second black ball given that
4
4
the first drawn is black and not replaced is: P (B/A) =
=
10+4
14
Since, the events are dependent, so the probability that
both balls drawn are black is:
5
4
2
x =
15
14
21
Question: Find the probability of drawing a king, a queen
and a knave in that order from a pack of cards in three
consecutive draws, the cards drawn not being
replaced.
4
Solution: The probability of drawing a king = P (A) =
52
The probability of drawing a queen after a king has
been drawn
4
P (B/A) =
51
The probability of drawing a knave after a king and a
queen have
been drawn
4
P (C/AB) =
50
P (ABC) =
4
52
x
4
51
x
4
50
=
8
16575
COMBINED USE OF ADDITION AND MULTIPLICATION
THEOREM
Question: A speaks truth in 80% cases, B in 90% cases. IN what percentage
of cases are they likely to contradict each other in stating the same fact.
Solution: Let P (A) and P (B) denote the probability that A and B speak the
truth. Then,
80
4
4
1
P (A) = 100 = 5
P (𝐴) = 1 – P (A) = 1 - 5 = 5
90
9
9
1
P (B) = 100 = 10
P (𝐵 ) = 1 – P (B) = 1 - 10 = 10
They will contradict each other only when one of them speaks the truth
and the other speaks a lie.
Thus, there are two possibilities:
(i) A Speaks the truth and B tells a lie
(ii) B Speaks the truth and A tells a lie.
4
1
4
P (A𝐵) = 5 x 10 = 50
1
9
9
P (𝐴 B)= 5 x 10 = 50
4
9
13
Required Probability = 50 + 50 = 50
Question: A bag contains 5 white and 3 red balls and four
balls are successively drawn out and not replaced. What
is the chance that they are alternatively of different
colours?
Solution: 4 balls of alternative colours can be either White,
Red, White, Red or Red, White, Red, White
Beginning with White Ball:
5
The probability of drawing a white ball =
The probability of drawing a red ball =
3
7
8
The probability of drawing a white ball =
The probability of drawing a red ball =
5
8
3
7
4
6
2
5
P (1W 1R 1W 1R) = x x x =
1
14
2
5
4
6
Beginning with Red Ball:
3
The probability of drawing a red ball =
8
The probability of drawing a white ball =
The probability of drawing a red ball =
2
6
The probability of drawing a white ball =
3
8
5
7
2
6
4
5
P (1R 1 W 1R 1W) = x x x =
Required Probability
=
1
14
+
1
14
=
1
7
1
14
5
7
4
5
Permutation
 A Permutation is an arrangement in a definite order of n objects
taken r at a time.
nP = n (n-1) (n-2)….(n-r+1)
r
𝑛!
𝑛−𝑟 !
=
0 < r ≤ n and n! = n (n-1) (n-2)………3.2.1
Where repetition is not allowed.
nP = 𝑛𝑟 where repetition is allowed.
r
 The number of permutation of n objects where p objects are of
same kind and rest are different
𝑛!
𝑝!
=
 The number of permutation of n objects where 𝑝1 objects are of
one kind, 𝑝2 objects are of one kind….. 𝑝𝑘 objects are of one kind
and rest if any are different
𝑛!
1 !.𝑝2 ! ……..𝑝𝐾 !
=𝑝
 Find the number of permutations of the letters of the word
ALLHABAD.
Here Total no of Letters = 9
A’s or 𝑝1 = 4
L’s or 𝑝2 = 2
Required No. =
𝑛!
𝑝1 !.𝑝2 !
=
9!
4!.2!
=
9×8×7×6×5×4! 9×8×7×6×5
=
= 7560
4!.2×1
2
How many 4-digit numbers can be formed by using the digits
1 to 9 if (1) repetition is allowed (2) repetition is not allowed.
when repetition is allowed nPr = 𝑛𝑟 = 94 = 9×9×9×9= 6561
when repetition is not allowed nPr =
𝑛!
𝑛−𝑟 !
=
9!
9−4 !
9!
5!
= =3024
 Find the number of permutations of the letters of the word
ALLHABAD.
Here Total no of Letters = 9
A’s or 𝑝1 = 4
L’s or 𝑝2 = 2
Required No. =
𝑛!
𝑝1 !.𝑝2 !
=
9!
4!.2!
=
9×8×7×6×5×4! 9×8×7×6×5
=
= 7560
4!.2×1
2
How many 4-digit numbers can be formed by using the digits
1 to 9 if (1) repetition is allowed (2) repetition is not allowed.
when repetition is allowed nPr = 𝑛𝑟 = 94 = 9×9×9×9= 6561
when repetition is not allowed nPr =
𝑛!
𝑛−𝑟 !
=
9!
9−4 !
9!
5!
= =3024
Combination
A Combination is an arrangement of n objects taken r
at a time.
nC = n (n-1) (n-2)….(n-r+1)/ r!
r
=
𝑛!
𝑟! . 𝑛−𝑟 !
0 < r ≤ n and n! = n (n-1) (n-2)………3.2.1
Note : Order is not important in case of combination
Example
 A committee of 3 persons is to be constituted from a group of
2 men and 3 women. What is the probability that the
committee would consist of 1 man and two women.
 Total Persons = 2+3 = 5
3 persons can be chosen out of 5 =
5C
3
1 man can be chosen out of 2 = 2C1 =
=
2!
1! . 2−1 !
2 Women can be chosen out of 3 = 3C2 =
Required Probability =
2×3
10
=
6
3
=
10 5
5!
3! . 5−3 !
= 10 ways
= 2 ways
3!
2! . 3−2 !
= 3 ways
BERNOULLI’S THEOREM IN THE THEORY OF
PROBABILITY
Bernoulli’s theorem is very useful in working out various probability problems. This theorem
states that if the probability of happening of an event on one trial or experiment is known,
then the probability of its happening exactly, 1,2,3,…r times in n trials can be determined by
using the formula:
P (r) = nCr pr . qn-r
r = 1,2,3,…n
Where,
P (r) = Probability of r successes in n trials.
p = Probability of success or happening of an event in one trial.
q = Probability of failure or not happening of the event in one trial.
n = Total number of trials.
Example
 A Box of 9 golf gloves contains 2 left-handed and 7 right
handed. If 2 gloves are randomly selected that both gloves are
of (i) of same hand (ii) of right hand (iii) of both hand.
2C
7C
42 44 11
 (i) P(LL or RR) = 9 + 9 =
+
= =
72 18
C2 C2 36 72
7C
42 7
 (ii) P(RR)= 9 2 =
=
C2 72 12
2C1
7C
7C
2C 1
2 7
7 2
1
1
 (iii) P (Both Hands) =( 9
× 9 ) + (9 × 9 ) =( × )+ ( × )
9 8
9 8
C1
C1
C1
C1
7
=
2
18
2
1
EXAMPLES ILLUSTRATING THE APPLICATION OF BERNOULLI’S
THEOREM
Question: Three coins are tossed simultaneously. What is the probability that there will be
exactly two heads?
Solution: P (r) = = nCrpr . qn-r
1
Given, n = 3, r = 2, p = probability of head in throw of a coin = 2
1
2
q=1- =
1
2
1
2
1
2
P (2H) = 3C2 ( ) 2 . ( )3-2
=
3!
1
x =
3 −2 ! 2! 8
=
3
8
3x
1
8
BAYE’S THEOREM
Bayes' Theorem is named after the British Mathematician Thomas Bayes and it was published
in the year 1763. With the help of Bayes' Theorem, prior probability are revised in the light of
some sample information and posterior probabilities are obtained. This theorem is also called
Theorem of Inverse Probability.
STATEMENT OF BAYE’S THEOREM
If A1 and A2 are mutually exclusive and exhaustive events and B be an event which can occur
in combination with A1 and A2, then the conditional probability for event A1 and A2 given the
event B is given by:
𝐵
𝑃 𝐴1 𝑃 𝐴
.
1
P (A1/B) =
𝐵
𝐵
𝑃 𝐴1 𝑃 𝐴 +𝑃 𝐴2 𝑃 𝐴
.
.
1
2
Similarly,
𝐵
𝑃 𝐴2 𝑃 𝐴
.
2
P (A2/B) =
𝐵
𝐵
𝑃 𝐴1 𝑃 𝐴 +𝑃 𝐴2 𝑃 𝐴
.
.
1
2
PROOF OF THE THEOREM
A1
A2
B
Since, A1 and A2 are mutually exclusive events and since the event B occurs with only one of
them, so that
B = BA1 + BA2
or
B = A1B + A2B
By the addition theorem of probability, we have
P (B) = P (A1B) + P(A2B)
…(i)
Now, by the multiplication theorem, we have
P (A1B) = P (A1) . P (B/A1)
…(ii)
P (A2B) = P (A2) . P (B/A2)
…(iii)
Substituting the values of P (A1B) and P (A2B) in (i), we get
P (B) = P (A1) . P (B/A1) + P (A2) . P (B/A2)
…(iv)
Hence,
P (B) = 2𝑖=1 P (Ai) . P (B/Ai)
…(v)
Again by the theorem of conditional probability, we have
𝑃 (𝐴1𝐵)
P (A1/B) =
…(vi)
𝑃 (𝐵)
Substituting the values of P (A1B) and P (B) from (ii) and (iv) in equation (vi), we get
P (A1/B) =
Similarly,
P (A2/B) =
𝐵
𝐴1
𝑃 𝐴1 . 𝑃 ( )
𝐵
𝐵
𝑃 𝐴1 . 𝑃 𝐴 +𝑃 𝐴2 .𝑃 (𝐴 )
1
2
𝐵
𝐴2
𝑃 𝐴2 . 𝑃 ( )
𝑃 𝐴1 . 𝑃
𝐵
𝐴1
𝐵
𝐴2
+𝑃 𝐴2 .𝑃 ( )
The probabilities P (A1) and P (A2) are called posterior probabilities and probabilities P (A1/B)
and P (A2/B) are called posterior probabilities.
The theorem can be expressed by means of the following figure:
P (B/A1)
P (A1) . P (B/A1)
First Branch
P (A1)
P (A2)
Second Branch
P (B/A2)
Prior Probability
Of A1 and A2
Conditional Probability
of B given A1 and A2
P (A2) . P (B/A2)
Joint Probability
Now,
P (A1/B) =
=
𝐽𝑜𝑖𝑛𝑡 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑏𝑟𝑎𝑛𝑐ℎ
𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑗𝑜𝑖𝑛𝑡 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑏𝑟𝑎𝑛𝑐ℎ𝑒𝑠
𝐵
𝐴1
𝑃 𝐴1 .𝑃 ( )
𝐵
𝐵
𝑃 𝐴1 . 𝑃 𝐴 +𝑃 𝐴2 . 𝑃 (𝐴 )
1
2
Similarly,
𝐽𝑜𝑖𝑛𝑡 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 2𝑛𝑑 𝑏𝑟𝑎𝑛𝑐ℎ
P (A2/B) =
𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑗𝑜𝑖𝑛𝑡 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑏𝑟𝑎𝑛𝑐ℎ𝑒𝑠
=
𝐵
𝐴2
𝑃 𝐴2 .𝑃 ( )
𝐵
𝐵
𝑃 𝐴1 .𝑃 𝐴 +𝑃 𝐴2 .𝑃 (𝐴 )
1
2
EXAMPLES ILLUSTRATING THE APPLICATION OF BAYE’S
THEOREM
Question: In a bolt factory machine A, B and C manufacture respectively 25%, 35% and 40%
of the total. Of their output 5, 4, 2 per cent are defective bolts. A bolt is drawn at
random from the product and is fount to be defective. What is the probability that
it was manufactured by machine C?
Solution: Let A, B and C be the events of drawing a bolt produced by machine A, B and C
respectively and let D be the event that the bolt is defective.
The prior probabilities are:
The conditional probabilities are:
25
5
P (A) = 25% =
= 0.25
P (D/A) = 5% =
= 0.05
100
100
35
4
P (B) = 35% =
= 0.35
P (D/B) = 4% = 100 = 0.04
100
40
2
P (C) = 40% = 100 = 0.40
P (D/C) = 2% = 100 = 0.02
Events
(1)
Prior Probability
(2)
Conditional
Probability (3)
Joint Probability
(2) X (3)
A
P(A) = 0.25
P (D/A) = 0.05
0.25x0.05
B
P (B) = 0.35
P (D/B) = 0.04
0.35x0.04
C
P (C) = 0.40
P (D/C) = 0.02
0.40x0.02
𝐽𝑜𝑖𝑛𝑡 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑐ℎ𝑖𝑛𝑒 𝐶
P (C/D) =
𝑆𝑢𝑚 𝑜𝑓 𝐽𝑜𝑖𝑛𝑡 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑇ℎ𝑟𝑒𝑒 𝑀𝑎𝑐ℎ𝑖𝑛𝑒𝑠
=
0.40 𝑋 0.02
0.25 𝑋 0.05+0.35 𝑋 0.04+0.40 𝑋 0.02
0.008
0.008
=
=
0.0125+0.014+0.008 0.0345
= 0.2318 or 23.18%
Example
An aircraft emergency locator transmitter (ELT) is a device
designed to transmit a signal in the case of a crash. The Altigauge
Manufacturing Company makes 80% of the ELTs, the Bryant
Company makes 15% of them, and the Chartair Company makes
the other 5%. The ELTs made by Altigauge have a 4% rate of
defects, the Bryant ELTs have a 6% rate of defects, and the
Chartair ELTs have a 9% rate of defects (which helps to explain
why Chartair has the lowest market share).
a. If an ELT is randomly selected from the general population of all
ELTs, find the probability that it was made by the Altigauge
Manufacturing Company.
b. If a randomly selected ELT is then tested and is found to be
defective, find the probability that it was made by the Altigauge
Manufacturing Company.
 A factory has two machines A and B. Past records show that
machine A produces 30% of the total output and machine B
70%. Machine A produces 5% defective articles and machine B
produces 1% defective items. An item is drawn at random and
found to be defective. What is the probability that it was
produced by machine A.
More Examples
 A man is known to speak truth 3 out of 4 times. He throws a
dice and reports that it is a six. Find the probability that it is
actually six.
 A restaurant serves two special dishes, A and the rest B to its
customers consisting of 60 % men and 40% women. 80% of
men order dish A and the rest dish B. 70% of women order
dish B and the rest A. In what ratio of A to B should the
restaurant prepare the two dishes.
 There are two identical boxes containing respectively 4
white and 3 red balls; and 3 white and 7 red balls
respectively. A ball is drawn and it is white what is the
probability that is drawn from first box.