1 - Probabilty Introduction .ppt

MCA-105
Statistics and
Probability

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PART :II (Probability)
• PART :II (Probability)
1. Introduction to Probability
2. Conditional Probability
3. Random variables
4. Mathematical Expectation

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Probability
• Random experiment,
• sample space,
• events,
• axiomatic Probability,
• Algebra of events
• Conditional Probability
• Multiplication theorem of Probability
• Independent events
• System reliability & Baye’s Theorem

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Reference Books
1.Fundamental of Mathematical Statistics –
S.C.Gupta, V.K.Kapoor
2.Schaum’s Outlines Probability, Random
Variables & Random Process Tata McGraw Hill

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Introduction
• Probability is the chance that something will
happen
• Probability is a measure of how likely it is that
some event will occur

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Introduction
• If an experiment is repeated under essentially
homogeneous and similar conditions then two type of
situations are there:-
– The result is unique or certain
– The result is not unique but may be one of the several possible
outcomes
• The phenomena covered by situation 1 are known as
deterministic or predictable phenomena.
• The phenomena covered by situation 2 are known as
unpredictable or probabilistic phenomena.

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Example of Unpredictable or probabilistic phenomena
• In a toss of a uniform coin we are not sure of getting the head or tail
• A sales manager cannot predict with certainty about the sales target
next year
• Possibly, it will rain tomorrow
• There is a high chance of getting the distinction
• This year’s demand for the product is likely to exceed that of the last
year’s
• The expressions like ‘possibly’ , ‘high chance’ , ‘likely’ are indicates a
degree of uncertainty about the happening of the event
• There are three possible state of expectation
– Certainty (1) -- Impossibility (0)
– Uncertainty ( between 0 and 1)

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Basic terminology
Random Experiment
• It is an experiment, trial, or observation that can be repeated
numerous times under the same conditions
Outcome
• The result of a random experiment
Examples of a Random experiment include:
• The tossing of a coin. The experiment can yield two possible
outcomes, heads or tails
• The roll of a die. The experiment can yield six possible outcomes, this
outcome is the number 1 to 6 as the die faces are labeled
• The selection of a numbered ball (1-50) in a box. The experiment can
yield 50 possible outcomes

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Basic terminology
Sample Space
• The set of all possible outcomes is called as sample space.
It is denoted by S or Ω
• A particular outcome that is an element of S is called a
sample point
Examples
• The tossing of a coin, sample space is {Heads, Tails}
• The roll of a die, sample space is {1, 2, 3, 4, 5, 6}
• The selection of a numbered ball (1-50) in an urn, sample
space is {1, 2, 3, 4, 5, ...., 50}

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• The set of desired outcome is known as event
• An event is a subset of the sample space
• The event [a] consisting of a single point is called an
elementary event
• The empty set Ф and S are the subsets of S and hence
they are events
• Ф is called the impossible or null event
• S is called the certain or sure event
• Events can be combined to form new events:-
– A U B is the event iff A or B occurs ( or both)
– A ∩ B is the event iff A occurs and B occurs
– A’, the complement of A, is the event that occurs iff A does not
occur
Event

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Basic terminology
Example of Event
Experiment : Toss a die and observe the number that
appears on the top face
Sample Space S = { 1,2,3,4,5,6}
Let A, B, C are the three events where
A=An even number occurs B=An odd number occurs
A={2,4,6} B={1,3,5}
C= Number greater than three occurs
C= {4,5,6}
Then find A U C, A ∩ C , C’

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• Write the sample space for the given experiment
– A single coin is tossed
– Two coins are tossed
– A die is rolled
– A die is rolled and a coin is tossed
– Two dice are rolled
– Three coins are tossed
– A card drawn at random from a deck of cards
Do it Yourself

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Examples
– A single coin is tossed – {T,H}
– Two coins are tossed – {(T,T),(T,H),(H,T),(H,H)}
– A die is rolled – {1,2,3,4,5,6}
– A die is rolled and a coin is tossed -
{(1,H)(2,H)(3,H)(4,H)(5,H)(6,H)(1,T)(2,T)(3,T)(4,T)(5,T)(6,T)}
– Two dice are rolled –
{(1,1)(1,2)(1,3)(1,4)(1,6)…..(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}
– Three coins are tossed –
{(TTT)(TTH)(THH)(THT)(HHT)(HTH)(HHH)…}
– A card drawn at random from a deck of cards {1,2….52}

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The Basic Principle of Counting
• Suppose that two experiments are to be
performed.
• Then if experiment 1 can result in any one of m
possible outcomes and if for each outcome of
experiment 1 there are n possible outcomes of
experiments 2, then together there are mn
possible outcomes of the two experiments

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Example
• A Small Community consists of 10 women, each of
whom has 3 children. If one woman and one of her
children are to be chosen as mother and child of the
year, how many different choices are possible?
Solution:
By regarding the choice of the woman as the outcome of
the first experiment and the subsequence choice of the
one of her children as the outcome of second experiment,
so as per the principle 10*3 = 30 possible choices

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16
Permutations vs. Combinations
• Both are ways to count the possibilities
• The difference between them is whether order
matters or not
• Consider a poker hand:
– A♦, 5♥, 7♣, 10♠, K♠
• Is that the same hand as:
– K♠, 10♠, 7♣, 5♥, A♦
• Does the order the cards are handed out matter?
– If yes, then we are dealing with permutations
– If no, then we are dealing with combinations

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17
Permutations
• A permutation is an ordered arrangement of the elements
of some set S
– Let S = {a, b, c}
– c, b, a is a permutation of S
– b, c, a is a different permutation of S
• An r-permutation is an ordered arrangement of r elements
of the set
– A♦, 5♥, 7♣, 10♠, K♠ is a 5-permutation of the set of
cards
• The notation for the number of r-permutations: P(n,r)
– The poker hand is one of P(52,5) permutations

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Permutations
• r-permutation notation: P(n,r)
– The poker hand is one of P(52,5) permutations
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Sample question
• How many permutations of {a, b, c, d, e, f, g} end
with a?
Solution:
- Note that the set has 7 elements
The last character must be a
The rest can be in any order
Thus, we want a 6-permutation on the set {b, c, d, e,
f, g}
•P(6,6) = 6! = 720
•Why is it not P(7,6)?

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Combinations
• What if order doesn’t matter?
• In poker, the following two hands are equivalent:
– A♦, 5♥, 7♣, 10♠, K♠
– K♠, 10♠, 7♣, 5♥, A♦
• The number of r-combinations of a set with n
elements, where n is non-negative and 0≤r≤n is:
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A combination is a grouping of items in which order
does not matter. There are generally fewer ways to
select items when order does not matter. For
example, there are 6 ways to order 3 items, but they
are all the same combination:
6 permutations  {ABC, ACB, BAC, BCA, CAB, CBA}
1 combination  {ABC}

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Finding Permutations
How many ways can a student government select a
president, vice president, secretary, and treasurer from a
group of 6 people?
This is the equivalent of selecting and arranging 4 items from
6.
= 6 • 5 • 4 • 3 = 360
Divide out common factors.
There are 360 ways to select the 4 people.
Substitute 6 for n and 4 for r in

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Example Finding Permutations
How many ways can a stylist arrange 5 of 8 vases
from left to right in a store display?
Divide out common
factors.
= 8 • 7 • 6 • 5 • 4
= 6720
There are 6720 ways that the vases can be arranged.

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Awards are given out at a costume party. How
many ways can “most creative,” “silliest,” and
“best” costume be awarded to 8 contestants if no
one gets more than one award?
= 8 • 7 • 6
= 336
There are 336 ways to arrange the
awards.
Do it Yourself

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How many ways can a 2-digit number be formed by using
only the digits 5–9 and by each digit being used only
once?
= 5 • 4
= 20
There are 20 ways for the numbers to be formed.
Do it Yourself

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Application
There are 12 different-colored cubes in a bag. How many
ways can randomly draw a set of 4 cubes from the bag?
Step 1 Determine whether the problem represents
a permutation of combination.
The order does not matter. The cubes may
be drawn in any order. It is a combination.

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Example Continued
= 495
Divide out
common
factors.
There are 495 ways to draw 4 cubes from 12.
5
Step 2 Use the formula for combinations.
n = 12 and r = 4

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Check It Out!
The swim team has 8 swimmers. Two swimmers
will be selected to swim in the first heat. How
many ways can the swimmers be selected?
= 28
The swimmers can be selected in 28 ways.
4
Divide
out
common
factors.
n = 8 and r = 2

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1. Six different books will be displayed in the library window.
How many different arrangements are there?
2. The code for a lock consists of 5 digits. The last number
cannot be 0 or 1. How many different codes are
possible? 80,000
720
3. The three best essays in a contest will receive gold, silver, and
bronze stars. There are 10 essays. In how many ways can the
prizes be awarded?
4. In a talent show, the top 3 performers of 15 will advance to the
next round. In how many ways can this be done? 2,730
720
Do it Yourself

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Probability
• If S is the sample space and A is an event then the
probability of Event A is defined by n(A) / n(S).
• It is denoted by P(A)
• P(A) = n(A) / n(S)

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Classical Probability Formula
• 0 ≤ P(A) ≤ 1 as 0 < m < n
• If P(A) = 0 then A is called a null event, or impossible event.
• If P(A) = 1 then A is called a sure event.
• If m is the number of cases favorable to A. Then
n - m is favorable to "non occurrence of A".
P(A’) = (n – m) / n = 1 – m/n = 1 – P(A)

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Statistical or Empirical Probability
• If an experiment is repeated N number of times under essential
homogeneous and identical conditions and an event E happens M
times, then the probability of happening of E is given by :-
P(E) = lim ( M / N )
N→∞
• A pair of dice is rolled. Find the probability that
– Both the dice show the same number
– The first die shows 6
– The total of the numbers on the dice is 8
– The total of the numbers on the dice is greater than 8
– The total of the numbers on the dice is 13
– The total of the numbers on the dice is any number from 2 to 12
both inclusive

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Example
• A pair of dice is rolled. Find the probability that
– Both the dice show the same number Ans : 6/36
– The first die shows 6 Ans : 6/36
– The total of the numbers on the dice is 8 Ans : 5/36
– The total of the numbers on the dice is greater than 8 Ans : 7/36
– The total of the numbers on the dice is 13 Ans : 0
– The total of the numbers on the dice is any number from 2 to 12
both inclusive Ans : 1

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Basic concept of drawing a card
• In a pack or deck of 52 playing cards, they are
divided into 4 suits of 13 cards each i.e. spades ♠
hearts ♥, diamonds ♦, clubs ♣.
• Cards of Spades ♠ and clubs ♣ are black cards.
• Cards of hearts ♥ and diamonds ♦ are red cards.
• The card in each suit, are ace, king, queen, jack or
knaves, 10, 9, 8, 7, 6, 5, 4, 3 and 2.
• King, Queen and Jack are face cards. So, there are
12 face cards in the deck of 52 playing cards.
•

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Examples
• A card is drawn from a deck. Find the probability that a
card is a :
– King
– An ace
– A club
– A red card
– A face card
– A red face card
• What is the chance that a leap year selected at random will
contain 53 Sundays?

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A card is drawn from a deck. Find the probability that a card is a :
1) King 2) An ace 3) A club 4) A red card 5) A face card 6) A red
face card
1) A king
Total number of king is 4 out of 52 cards.
Therefore, probability of getting ‘a king’
Number of favorable outcomes
P(E) = Total number of possible outcome
= 4/52
= 1/13
2) A Ace
Total number of king is 4 out of 52 cards.
= 4/52
= 1/13

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face card
3) A Club
Total number of club is 13 out of 52 cards.
= 13/52
= 1/4
4) A Red Card
Total number of red card is 26 out of 52 cards.
= 26/52
=1/2

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face card
5) A face card
Total number of face card is 12 out of 52 cards.
= 12/52
= 3/13
6) A Red Face Card
Total number of red face card is 6 out of 52 cards.
= 6/52
= 3/26

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What is the chance that a leap year selected at
random will contain 53 Sundays?
Solution:
In a leap year there are 52 weeks and 2 days which
can be any two days
Sample Space:
(Sunday ,Monday), (Monday, Tuesday)
(Tuesday, Wednesday), (Wednesday, Thursday)
(Thursday, Friday), (Friday, Saturday)
(Saturday, Sunday)
i.e. There are 7 possibilities out of which 2
possibilities contain Sunday.
Possible Event is 2
Required probability = 2/7

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Probability
• Product Rule: Suppose a procedure can be broken down
into a sequence of two tasks. If there are n1 ways to do
the 1st task and n2 ways to do the 2nd task after the first
task has been done, then there are n1 * n2 ways to do the
procedure.
• Sum Rule: If a first task can be done in n1 ways and a
second task in n2 ways and if these tasks cannot be done
at the same time, then there are n1 + n2 ways to do either
task.

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Examples
• A bag contains 3 red and 6 white and 7 blue balls. What is
the probability that one is white and other is blue?
• Box A contains 5 red marbles and 3 blue marbles and Box B
contains 3 red and 2 blue marbles. A marble is drawn at
random from each box. Find the probability that one is red
and one is blue?
• Two cards are drawn from a pack of 52 playing cards. Find
the probability that both the cards are aces.
• A box contains 2 white socks and 2 blue socks. Two are
drawn at random. Find the probability that they match.
• 5 horses are in race. Sumit picks 2 of the horses at random
and bets on them. Find the probability that Sumit picked
the winner.

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A bag contains 3 red and 6 white and 7 blue balls. What is the
probability that one is white and other is blue?
Solution:
• Let S be the sample space.
• Then, n(S) = number of ways of drawing 2 balls out
of 15 = 16C2 = (16 x 15 x 14!) /(2 x 1 x 14!)= 120
• Let E = event of getting one while and one blue ball
• n(E) = 6C1*
7C1 = (6 x 7) = 42
• P(E) = n(E) /n(S)= 42/120 =7/20

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Box A contains 5 red marbles and 3 blue marbles and Box B contains 3
red and 2 blue marbles. A marble is drawn at random from each box.
Find the probability that one is red and one is blue?

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Two cards are drawn from a pack of 52 playing cards. Find the
probability that both the cards are aces.
Solution:
• hen, n(S) = number of ways of drawing 2 cards out
of 52 = 52C2
• Let E = event of getting two aces
• n(E) = 4C2
• P(E) = n(E) /n(S)= 4C2 / 52C2 = 1/221

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A box contains 2 white socks and 2 blue socks. Two are drawn
at random. Find the probability that they match.
Solution:
• There are c(4,2) = 4C2 = 6 ways to draw 2 of the
socks
• Then, n(S) = 6
• Let E = event of getting two socks with same color
• n(E) = 2
• P(E) = n(E) /n(S)= 2/6

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5 horses are in race. Sumit picks 2 of the horses at random
and bets on them. Find the probability that Sumit picked the
winner.
Solution:
• There are c(5,2) = 5C2 = 10 ways to draw 2 of the
hourses out of 5
• Then, n(S) = 10
• Four of the pairs will contain the winner e.g. h1 is
the winner than {h1h2,h1h3,h1h4,h1h5}
• Let E = event of getting the winner in set of two
hourses
• n(E) = 4
• P(E) = n(E) /n(S)= 4/10 = 2/5

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Examples
• Four cards are drawn at random from a pack of 52 cards.
Find the probability that
• They are a king, a queen, a jack and an ace
• Two are kings and two are queens
• Two are black and two are red
• There are two cards of hearts and two cards of diamonds
• There is one card of each suit
• A committee of 4 people is to be appointed from 3 officers
of the production department, 4 officers of the purchase
department, 2 officers of the sales department and 1
chartered accountant. Find the probability of forming the
committee in the following manner:
– There must be one from each category
– It should have at least one from the purchase department
– The chartered accountant must be in the committee

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Four cards are drawn at random from a pack of 52 cards. Find
the probability that
i. They are a king, a queen, a jack and an ace
ii. Two are kings and two are queens
iii. Two are black and two are red
iv. There are two cards of hearts and two cards of diamonds
v. There is one card of each suit

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A committee of 4 people is to be appointed from 3 officers of the
production department, 4 officers of the purchase department, 2
officers of the sales department and 1 chartered accountant. Find the
probability of forming the committee in the following manner:
- There must be one from each category
- It should have at least one from the purchase department
- The chartered accountant must be in the committee

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• An urn contains 6 white, 4 red and 9 black balls. If three balls are
drawn at random, find the probability that
• Two of the balls drawn are white
• One is of each color
• None is red
• At least one is white
• Four cards are drawn from a pack of 52 playing cards. Find the
probability that:
– All are diamond cards
– There is one card of each suit
– 2 spade and 2 hearts
• A box with 15 integrated chips contains 5 defective. If random
samples of 3 chips are drawn, what is the probability that all three
are defective?
Do it Yourself

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An urn contains 6 white, 4 red and 9 black balls. If three balls
are drawn at random, find the probability that
- Two of the balls drawn are white
- One is of each color
- None is red
- At least one is white

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Four cards are drawn from a pack of 52 playing cards. Find
the probability that:
- All are diamond cards
- There is one card of each suit
- 2 spade and 2 hearts
I ) 13C4
52C4
2 ) 13C1 x 13C1 x 13C1 x 13C1
52C4
3 ) 13C2 x 13C2
52C4

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A box with 15 integrated chips contains 5 defective. If
random samples of 3 chips are drawn, what is the probability
that all three are defective?
1 ) 5C3
15C3

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• What is the probability that 4 S’s come consecutively in
‘MISSISSIPPI’?
• How many even 2-digit numbers can be constructed out of digits 3,
4, 5, 6 and 7. Assume that i) you may use the same digit again ii)
you may not use same digit again.
• A box has 75 good IC chips and 25 defective chips. If 12 IC are
selected at random, find the probability that at least 1 chip is
defective?
• A room has 3 lamp sockets for which 3 bulbs are chosen from a
group of 6 working and 4 non-working bulbs. What is the
probability that the room is lit?
Example

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What is the probability that 4 I’s come consecutively in
‘MISSISSIPPI’?
Number of Letters/Characters in the word "MISSISSIPPI" = 11 {M, I, S, S, I, S,
S, I, P, P, I} ⇒ nL = 11
No. of Letters : a First Kind ⇒ No. of I's = 4 ⇒ a = 4
a Second Kind ⇒ No. of S's = 4 ⇒ b = 4
a Third Kind ⇒ No. of P's = 2 ⇒ c = 2
which are all different = 1 {M} ⇒ x = 1 nL = a + b + c + x
In the experiment of testing for the number of words that can be formed using
the letters of the word "MISSISSIPPI"
Total No. of Possible Choices = Number of words that can be formed using the
11 letters of the word "MISSISSIPPI" ⇒ n = nL / a! × b! × c!
= 11! / 4! × 4! × 2! = 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4! / 4! × 4 × 3 × 2 × 1 × 2 × 1 =
11 × 10 × 9 × 7 × 5 = 34,650
Sample Space n(S) = 34650

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What is the probability that 4 S’s come consecutively in
‘MISSISSIPPI’?
in total ⇒ nL(s) = 8 {M,I, I, I, I, (S, S, S, S), P, P}
of the first kind ⇒ No. of I's = 4 ⇒ p = 4
of the second kind ⇒ No. of P's = 2 ⇒ q = 2
which are all different = 2 {M, (S, S, S, S)} ⇒ y = 2 nL(s) = p + q + y
Number of Favourable/Favorable Choices = The number of words that can be
formed using the letters of the
word "MISSISSIPPI" such that all the I's are together ⇒ mA = (No. of ways in
which the 8 letters (taking the 4 “S's" as a unit are
can be arranged in the 8 places)
× (No. of ways in which the 4 “S's" can be inter arranged
between themselves) = (nL(s))!p! × q! × 4!4! = 8! / 4! × 2! × 1 = 8 × 7 × 6 × 5
4! /4! × 2 × 1 = 8 × 7 × 3 × 5 = 840
P(E) = N(E) / N(S) = 840/34650

XP
How many 2-digit numbers can be constructed out of digits 3,
4, 5, 6 and 7. Assume that i) you may use the same digit again
ii) you may not use same digit again.
1. Can use the same digit again
Total digit 3,4,5,6 and 7 = 5 digits
so, total 5C1*5C1= 5*5 = 25 ways we can select 2 digit
numbers
2. Can not use the same digit again
Again Total digit 3,4,5,6 and 7 = 5 digits
so, for first digit choices are 5 digits and for second digit
choices left are 4 digits
so, 5C1*4C1 = 5*4 = 20 different digits can be form

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• A room has 3 lamp sockets for which 3 bulbs are chosen
from a group of 6 working and 4 non-working bulbs. What
is the probability that the room is lit?
Solution:
• Let S be the sample space.
• Then, P(room is lit) = 1-P(room is dark)
• P(Room is dark) = 4C3/10C3
= 4/120
= 1-1/30
= 29/30

XP
• In a random arrangement of the letters of the word
‘COMMERCE’, find the probability that all the vowels
come together. 3/28
• What is the probability that at least 2 out of 5 people
have the same birthday, assume a 365 days in a year and
all days are equally likely.
• In a group of 4 IC chips containing 2 good and 2
defective chips, if 3 chips are selected at random, what
is the probability that 2 of 3 selected are defective. 1/2
Do it Yourself

XP
Examples
• N persons are seated on n chairs at a round table. Find
the probability that two specified persons are sitting
next to each other?
• If n biscuits are distributed among N beggars, find the
probability that a particular beggar gets r biscuits such
that r < n.
• A series of n jobs arrive at a computing center with n
processors. Assume that each of the nn possible
assignment vectors (processor for job 1, ….., processor
for job n) is equally likely. Find the probability that
exactly one processor will be idle?

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n person seated n chair (n-1)! Ways.
Assuming specified person sitting togather as one,
For example A and B
Than (n-2)! Ways for the arrangement
Plus two person can interchange the position
So, (n-2)!2! Ways total for the arrangement
Probability of two person sitting next to each other = n(a)/n(s)
= 2! * (n-2)!
(n-1)!
= 2/(n-1)
N persons are seated on n chairs at a round table. Find the
probability that two specified persons are sitting next to each
other?
A
B
C

XP
If n biscuits are distributed among N beggars, find the
probability that a particular beggar gets r biscuits such that r
< n.
n biscuits can be distributed among N beggars in N n ways:
1st-------any of the N beggar
2nd---------any of the N beggars
nth------------any of the N beggars
n(s) = N *N*N…..(n times) = Nn
Moreover, r biscuits can be given to particular beggar in n C r
ways. The remaining (n-r) biscuits will be distributed among the
remaining (N-1) beggars in (N-1) n-r ways. Thus, the total number
of favorable ways :
n(A)= n C r (N-1) n-r
Hence the required Probability = n(A) /n(S)
= n C r (N-1) n-r
N n

XP
A series of n jobs arrive at a computing center with n processors.
Assume that each of the nn possible assignment vectors (processor for
job 1, ….., processor for job n) is equally likely. Find the probability that
exactly one processor will be idle? (without replacement)
Sample Space S= n jobs send to n processor for execution
n(S) = [ For 1st job we have 1…n choice,
2nd job we have 1…n choice……..
nth job we have 1….n choice ]
= n n
Assuming one processor one job and remaining n-2 has also one job and one is
idle
Event A = One processor 2 jobs and (n-1) processor remaining ( n-2) jobs
Choice of 2 job to 1 processor
nC2 + nC2 + nC2 + nC2 …….n terms
Any one processor out of n is selected and 2 jobs from n jobs are given to it….
So, 1st job has (n-1) choice
2nd job has (n-2) choice……………
so, [(n-1) (n-2) (n-3) (n-4)…2]
[nC2 + nC2 + nC2 ……n terms] (n-1) !
P(A) = nC2 *n * (n-1)!
n n
P(A) = nC2 * n!
n n

XP
MUTUALLY EXCLUSIVE EVENTS
• Addition Theorem of probability: If A and B are two
events of a sample space S then
P (A U B) = P (A) + P (B) – P (A ∩ B)
• Mutually Exclusive Events:
– Events are said to be mutually exclusive events if the happening
of any one of them stops the happening of all the others
– No two or more events can happen simultaneously in the same
trial
– If A and B are two events of a sample space S such that A ∩ B = ø
then A and B are said to be mutually exclusive events. If A and B
are mutually exclusive events then
P (A U B) = P (A) + P (B)

XP
Important Theorms
• P(A’) = 1-P(A)
• P(A∩B) = P(A) +P(B) – P(AUB)
• P(A’UB’) = P(A’)+P(B’)-P(A’∩B’)
• or = 1-P(A ∩B)
• P(A ∩B’) = P(A) – P(A∩B)
• P(A’ ∩ B’) = 1-P(AUB)
• P(A ∩ B’ ∩ C’) =P(A)- P(A ∩B)-P(A ∩C) +P(A ∩B ∩C)
• P(AUBUC)=P(A)+P(B)+P(C)- P(A ∩B) - P(A ∩C) –
P(B∩C) + P(A ∩ B ∩ C)

XP
Three Arbitrary events A, B and C Then probability of
i) Only A occur ii) Both A and B But not C iii) All three Events
Occurs iv) at least one occur v) At lease Two occur vi) one and
no more occurs vii) Two and no more occur viii) None occurs
i) A∩B’∩C’
ii) A∩B∩C’
iii) A∩B∩C
iv) AUBUC
v) (A∩B∩C’) U (A∩B’∩C)U (A’∩B∩C) U (A∩B∩C)
vi) (A∩B’∩C’) U (A’∩B∩C’)U (A’∩B’∩C)
vii) (A’∩B∩C) U (A∩B∩C’)U (A∩B’∩C)
viii)(A’∩B’∩C’)
A B
C
S

XP
• The probability of occurrence of an event A is 0.7, the
probability of occurrence of B is 0.4 and that of at least
one of A and B not occurring is 0.6. Find the probability
that at least A and B occurs
• Two dice are tossed. Find the probability of getting an
even number on the first die or a total of 8
• The probability that a certain film gets award for its
story is 0.23, it will get award for its music is 0.15 and it
will get award for both is 0.07. What is the probability
that film will get award for a) at least one of the two b)
exactly one of the two.

XP
The probability of occurrence of an event A is 0.7,
the probability of occurrence of B is 0.4 and that of
at least one of A and B not occurring is 0.6. Find the
probability that at least one of A and B occurs
P(A) = 0.7 P(A’) = 1- P(A) = 0.3;
P(B) = 0.4 P(B’) = 1-P(B)= 0.6;
P(A’ U B’) = 0.6
P(A U B) = ?
P(A’ U B’) = 1- P(A ∩ B)
P(A ∩ B) = 0.4
P(A UB) = P(A) +P(B) –P(A ∩ B)
= 0.7+0.4-0.4
= 0.7

XP
Two dice are tossed. Find the probability of getting an
even number on the first die or a total of 8
S = {1,2,3,4,5,6} * {1,2,3,4,5,6} = 36
A = Even number = {2,4,6} * {1,2,3,4,5,6} = 18
P(A) = 18/36 = 1/2
B= Total of 8 = {(4,4),(5,3),(3,5),(2,6),(6,2)}
P(B) = 5/36
This is mutually exclusive events
P(AUB) = P(A) + P(B) – P(A ∩B)
= 18/36+5/36-3/36
=20/36 =5/9

XP
• An `MCA applies for two firms X and Y. The probability
of his being selected in firm X is 0.7 and being rejected
at Y is 0.5. The probability of at least one of his
application being rejected is 0.6. What is the probability
that he will be selected in one of the firms?
• A card is drawn from a deck of 52 cards. Find the
probability of getting a king or a heart or a red card
• The probability that a person stopping at a petrol pump
will ask for petrol is 0.8, will ask for water is 0.7 and for
both is 0.65. Find the probability that a person will ask
for a) either petrol or water b) neither petrol nor water
c) only water.

XP
An MCA applies for two firms X and Y. The probability of his
being selected in firm X is 0.7 and being rejected at Y is 0.5.
The probability of at least one of his application being
rejected is 0.6. What is the probability that he will be selected
in one of the firms?
P(X) =0.7
P(Y’) = 0.5  P(Y) =0.5
P(X’UY’) = 0.6
P(X’UY’) = 1-P(X ∩ Y)
0.6 = 1- P(X ∩ Y)
P(X ∩ Y) = 0.4
P(XUY) = P(X)+P(Y) - P(X ∩ Y)
= 0.7+0.5 -0.4 = 0.8

XP
A card is drawn from a deck of 52 cards. Find the
probability of getting a king or a heart or a red card
A= Event of getting King
P(A) = 4/52
B = Event of getting heart
P(B) = 13/52
C= Event of getting red card
P(C) = 26/52
P(AUBUC) = P(A)+P(B)+P(C) – P(A∩B) – P(B∩C)- P(A∩C) –P(A
∩B ∩C)
= 4/52+13/52+26/52 – 1/52 – 13/52 -2/52 +1/52
= 7/13

XP
A= Ask for Petrol
P(A) = 0.8
B= Ask for Water
P(B) = 0.7
P(A ∩B) = 0.65
The probability that a person stopping at a petrol
pump will ask for petrol is 0.8, will ask for water is 0.7
and for both is 0.65. Find the probability that a
person will ask for a) either petrol or water b) neither
petrol nor water c) only water.
a) P(AUB) = ?
P(AUB) = P(A) +P(B)- P(A∩B)
= 0.8+0.7-0.65
= o.85
b) P(A’∩B’) = 1 – P(A UB)
= 1 – 0.85
= 0.15
c) P(A’ ∩B) = P(B) – P(A ∩B)
= 0.7 – 0.65
= 0.05

XP
• Suppose A and B are events with P (A) = 0.6, P (B) = 0.3 and P (A ∩ B)
= 0.2. Find the probability that i) A does not occur, ii) B does not
occur, iii) A or B occurs, iv) neither A nor B occurs.
• The probability that a student passes a Physics test is 2/3 and the
probability that he passes both physics test and an English test is
14/45. The probability that he passes at least one test is 4/5. What is
the probability that he passes the English test?
• Sample survey was taken to check which newspaper people read (A,
B, C). In a sample of 100 people the following results are obtained. 60
read A, 40 read B, 70 read C, 45 read A and C, 32 read A and b, 38
read B and C, 30 read A, B and C. If a person is selected at random,
find the probability that a) he reads only A b) he reads at least two
newspapers c) he doesn’t read any paper.
Do it Yourself

XP
Probability Axioms
• Let S be a sample space ,A is the event and P(A) is
the probability measure associated with the event
A.
• The probability must satisfy the following axioms:
– For any event A, P(A) ≥ 0
– P(S) =1
– P( A U B) = P(A) + P(B),
where A and B are mutually exclusive events

XP
Conditional Probability
• If A and B are the events of sample space S,
then conditional probability of A given B is the
probability of A such that B has already
occurred and it is given by
P (A| B) = P (A ∩ B) / P (B)
• Multiplication Theorem of Probability: If A and
B are two events of a sample space S then
P (A ∩ B) = P (A| B) × P (B)

XP
Examples
• Let A and B be two events with P (A) = 3/8, P (B) = 5/8 and P (A U B) =
¾. Find P (A| B) and P (B| A).
• In a certain college, 25 percent of the students failed in mathematics,
15 percent failed in chemistry and 10 percent failed in both
mathematics and chemistry. A student is selected at random.
– If the student failed chemistry, what is the probability that he or she failed in
mathematics
– If the student failed mathematics, what is the probability that he or she failed in
chemistry
– What is the probability that the student failed in mathematics or chemistry.
– What is the probability that the student is failed neither in Mathematics nor in
Chemistry?
• From a city population, the probability of selecting (i) a male or a
smoker is 7/10 (ii) a male smoker is 2/5 and (iii) a male, if a smoker is
already selected is 2/3. Find the probability of selecting (a) a non-
smoker (b) a male (c) a smoker, if a male is first selected.

XP
Let A and B be two events with P (A) = 3/8, P (B) = 5/8
and P (A U B) = ¾. Find P (A| B) and P (B| A).
P(A ∩ B) = P(A) +P(B) – P(A UB)
= 3/8 +5/8 – 3/4
= 2/8
= 1/4
P(A|B) = n(A ∩ B) / n(B)
= 1/4 / 5/8
= 1 *8 / 4 *5
= 2/5
P(B|A) = n(A ∩ B) / n(A)
= 1/4 / 3/8
= 1 *8 / 4 *3
= 2/3

XP
In a certain college, 25 percent of the students failed in mathematics, 15
percent failed in chemistry and 10 percent failed in both mathematics
and chemistry. A student is selected at random.
i. If the student failed chemistry, what is the probability that he or she failed in
mathematics
ii. If the student failed mathematics, what is the probability that he or she failed in
chemistry
iii. What is the probability that the student failed in mathematics or chemistry.
iv. What is the probability that the student is failed neither in Mathematics nor in
Chemistry?
M = student fail in Maths C= Student Fail in the Chemistry
P(M) =0.25 P(C) = 0.15 P(M ∩ C) = 0.10
i) P(M|C) = P(M ∩C) /P(C) iv) P(M’ ∩ C’) = 1- P(MUC)
= 0.10 /0.15 = 2/3 =1-0.30
=0.70
ii) P(C|M) = P(M ∩C) /P(C)
= 0.10 /0.25 = 2/5
iii) P(MUC) = P(M)+P(C) – P(M ∩C)
= 0.25 +0.15 – 0.10 = 0.30

XP
From a city population, the probability of selecting (i) a male or a
smoker is 7/10 (ii) a male smoker is 2/5 and (iii) a male, if a smoker is
already selected is 2/3. Find the probability of selecting (a) a non-
smoker (b) a male (c) a smoker, if a male is first selected.
M = Male
S = Smoker
Given Data , P(MUS) = 7/10 P(M ∩ S) = 2/5 P(M|S) = 2/3
i) Non-smoker P(S’) =? iii) P(S|M) = P(M ∩ S) /P(M)
= 2/5 /1/2
P(M|S) = P(M ∩ S) /P(S) = 4/5
2/3 = 2/5 /P(S)
P(S) = 3/5
P(S’) = 1-P(S) = 2/5
ii) A male P(M) =?
P(MUS) =P(M)+P(S) – P(M ∩ S)
7/10 = P(M)+3/5-2/5
P(M) = 1/2

XP
Examples
• If A and B are two events and P(B) <1 prove that
P(A|B’) = {P(A) –P(A∩B)} / {1-P(B)}
• We are given a box containing 5000 IC chips, of which 1000 are
manufactured by company X and rest by company Y. 10% of the
chips made by company X and 5% of the chips made by company Y
are defective. If a randomly chosen chip is found to be defective,
find the probability that it comes from company X.
• Consider 4 computer firms A, B, C and D bidding for a contract. A
survey of past bidding success of this firm on similar contract gives
following probability of winning. P (A) = 0.35, P (B) = 0.15, P(C) =
0.3, P (D) = 0.2. Before the decision is made to avoid a contract,
firm B withdraws its bid. Find the new probabilities of winning the
bid for A, C and D.

XP
If A and B are two events and P(B) <1 prove that
P(A|B’) = {P(A) –P(A∩B)} / {1-P(B)}
.
Example 34

XP
We are given a box containing 5000 VLSI chips, 1000 of which are manufactured by
company X and the rest by company Y. Ten percent of the chips made by company X
are defective and 5% of the chips made by company Y are defective. If a randomly
chosen chip is found to be defective, find the probability that it came from company X.
Example 35

XP
Let A and B be events with P(A) =0.6, P(B) = 0.3 and P(A∩B) =0.2.Find P(A|B), P(B|A),
P(AUB) , P(A’),P(B’), P(A’|B’), P(B’|A’)
.
Do it yourself
i) P(A|B) = P(A ∩ B)/P(B) = 0.2/0.3 = 2/3
ii) P(B|A) = P(A ∩ B)/P(A) = 0.2/0.6 = 1/3
iii) P(AUB) = P(A)+P(B) - P(A ∩ B) = 0.6+0.3 – 0.2 = 0.7
iv) P(A’) = 1 – P(A) = 1-0.6 = 0.4
v) P(B’) = 1 – P(B) = 1-0.3 = 0.7
vi) P(A’|B’) = P(A’ ∩ B’)/P(B’) = 1- P(AU B) / 1-P(B) = 1-0.7 / 1-0.3 = 3/7
vii)P(B’|A’) = P(A’ ∩ B’)/P(A’) = 1- P(AU B) / 1-P(A) = 1-0.7 / 1-0.6 = 3/4

XP
Independent Events
• Two or more events are said to be independent if the
happening or non-happening of any one of them, does not
affect the happening of others
• If A and B are two events of a sample space S such that
P (A| B) = P (A)
then A and B are said to be independent events
• If A and B are independent events then
P (A ∩ B) = P (A) × P (B)

XP
Important points :-Independent Events
• If A and B are two mutually exclusive events, then A∩B =Ф , which
implies P(A∩B) =0. Now if they are independent then either P(A)
=0 or P(B) =0
• If the events A and B are independent and the events B and C are
independent ,then A and C need not be independent. In other
words, the relation of independence is not a transitive relation
• If A and B are independent events then so are events A’ and B,
events A and B’ and events A’ and B’

XP
Type of Independent Events
• Pairwise independent event
Two events are said to be pairwise independent event if
every pair of two events is independent
• Mutually independent event
The events in sample space S are said to be mutually
independent if the probability of the simultaneous
occurrence of finite number of them is equal to the
product of their separate probabilities

XP
Consider 4 computer firms A, B, C and D bidding for a contract. A survey
of past bidding success of this firm on similar contract gives following
probability of winning. P (A) = 0.35, P (B) = 0.15, P(C) = 0.3, P (D) = 0.2.
Before the decision is made to avoid a contract, firm B withdraws its bid.
Find the new probabilities of winning the bid for A, C and D.
Example 36

XP
Consider 4 computer firms A, B, C and D bidding for a contract. A survey of
past bidding success of this firm on similar contract gives following probability
of winning. P (A) = 0.35, P (B) = 0.15, P(C) = 0.3, P (D) = 0.2. Before the
decision is made to avoid a contract, firm B withdraws its bid. Find the new
probabilities of winning the bid for A, C and D.
Example :

XP
Example
• Two men A and B fire at a target. Suppose P(A) = 1/3 and P(B) =
1/5 denote their probabilities of hitting the target. Find the
probability that
– A does not hit the target
– Both hit the target
– One of them hit the target
– Neither hits the target
• Box A contains 5 red marbles and 3 blue marbles and Box B
contains 3 red and 2 blue. A marble is drawn at random from
each box.
– Find the probability p that both marbles are red
– Find the probability p that one is red and other is blue

XP
Two men A and B fire at a target. Suppose P(A) = 1/3 and P(B) = 1/5
denote their probabilities of hitting the target. Find the probability that
A does not hit the target
Both hit the target
One of them hit the target
Neither hits the target
Example 38

XP
Box A contains 5 red marbles and 3 blue marbles and Box B contains 3
red and 2 blue. A marble is drawn at random from each box.
Find the probability p that both marbles are red
Find the probability p that one is red and other is blue
Example 39

XP
EXAMPLE
• An electronic device is made up of 2 components A and B and is such
that it works as long as 1 component works. Probability of failure of A
is 0.2 and that of B is 0.1. If they work independently, find the
probability that device works.
• Prove that with example that mutual independence does not imply
pair wise independence.
• Prove that with example that any events may be pair wise
independent but need not to be mutually independent.
• If four squares are chosen at random on a chessboard, find the
chance that they should be in a diagonal line.

XP
An electronic device is made up of 2 components A and B and is such
that it works as long as 1 component works. Probability of failure of A is
0.2 and that of B is 0.1. If they work independently, find the probability
that device works.
Example 40
P(A’) = 0.2 P(B’) = 0.1
P(Device will work) = 1 - P(Device will not work)
= 1 – P(A’∩B’)
= 1- P(A’)P(B’)
= 1- 0.2 *0.1
= 0.998

XP
Pairwise Independence Example
• Suppose you toss a fair coin twice and let A be the event that the first
flip is H’s and B be the event that the second flip is H’s. Now let C be
the event that both flips are the same (i.e. both H’s or both T’s). Of
course A and B are independent.
• What is more interesting is that so are A and C: given that the first
toss came up H’s, there is still an even chance that the second flip is
the same as the first.
• Another way of saying this is that P[A∩C] = P[A]P[C] = 1/4 since A∩C
is the event that the first flip is H’s and the second is also H’s. By the
same reasoning B and C are also independent.
• On the other hand, A, B and C are not mutually independent. For
example if we are given that A and B occurred then the probability
that C occurs is 1. So even though A, B and C are not mutually
independent, every pair of them are independent. In other words, A,
B and C are pairwise independent but not mutually independent.

XP
• Consider the set of events A1, ..., An
• The events A1, ..., An are mutually independent if Pr[A1 ∩ A2... ∩
An] = Pr[A1]Pr[A2]...Pr[An]
• The events Ai and Aj are pairwise independent if Pr[Ai ∩ Aj] =
Pr[Ai]Pr[Aj] for all i, j
• Consider the set Ω = {abc, acb, aaa, bac, bca, bbb, cab, cba, ccc} and
probability of any of them being picked is 1/9
• Let event Ak denote that the kth letter is a. Clearly Pr[Ak ] = 1/3
• Pr[A1 ∩ A2] = Pr[A2 ∩ A3] = Pr[A3 ∩ A1] = 1/9
• Pr[A1 ∩ A2 ∩ A3] = 1/9 ≠ Pr[A1]Pr[A2]Pr[A3]
• Thus, the events pairwise independent but not mutually independent
• Pairwise Independence is weaker than Mutual Independence
Events pairwise independent but not mutually independent

XP
• A problem in Statistics is given to three students A,B and C
whose chances for solving it are ½, ¾, ¼ respectively. What
is the probability that the problem will be solved if all of
them try independently?
• It is 8:5 against the wife who is 40 years old living till she is
70 and 4:3 against her husband now 50 living till he is
80.Find the probability that:
– Both will be alive -- None will be alive
– Only wife will be alive --Only husband will be alive
– Only one will be alive --at least one will be alive
• Consider the following events for a family with children:
A={children of both genders}
B={at most one boy}
– Show that A and B are independent events if a family has 3
children
– Show that A and B are dependent events if a family has only 2
Do it Yourself

XP
A problem in Statistics is given to three students A,B and C whose
chances for solving it are ½, ¾, ¼ respectively. What is the probability
that the problem will be solved if all of them try independently?
Example 44
Probability that A failed to solve the problem = 1 – 1/2=1/2
Probability that B failed to solve the problem = 1 – 3/4 = 1/4
Probability that C failed to solve the problem = 1 – 1/4 = 3/4
Since the events are independent, the probability that all of them
failed to solve the
problem = ½ ¼ ¾ = 3/32
Hence the probability that the problem will be solved = 1 – 3/32=
29/32.

XP
It is 8:5 against the wife who is 40 years old living till she is 70 and 4:3
against her husband now 50 living till he is 80.Find the probability that:
1) Both will be alive 2) None will be alive
3) Only wife will be alive 4) Only husband will be alive
5) Only one will be alive 6) at least one will be alive
Example 45
Solution: Let A be the event that wife will be alive,
and B that husband will be alive, 30 years hence. Now, its given, P(A)
= 5/8+5 = 5/13 and P(B) = 3/4+3 = 3/7, this implies P (not A) = 1 – P(A)
= 8/13 and P (not B) = 4/7.
Here, A and B are independent and hence, all events, A, B, not A and not
B are also independent and hence,
1) P(A ∩ B)= P(A) * P(B)= 15/91 and
2) P(not A ∩ not B) = P(not A) * P(not B) = 32/91.
3) P(A ∩ not B) =P(A) – P (A ∩ B) = 5/13-15/91 = 20/91
4) P(not A ∩ B) =P(B) – P (A ∩ B) = 3/7-15/91 = 24/91
5) P(A not B) + P(not A ∩ B) = 20/91+24/91 = 44/91
6) P(A U B) = 1- P(not A ∩ not B) = 1-(32/91)=59/91

XP
Consider the following events for a family with children:
A={children of both genders}
B={at most one boy}
Show that A and B are independent events if a family has 3 children
Show that A and B are dependent events if a family has only 2 children
Example 46

XP
Law of Total Probability
• The law (or formula) of total probability is a
fundamental rule relating marginal probabilities to
conditional probabilities. It expresses the total
probability of an outcome which can be realized via
several distinct events - hence the name
• If B1, B2, B3, ….., Bn are mutually exclusive and
exhaustive events of the sample space S, then for any
event A of S.

XP
Three boxes contain red and green balls. Box 1 has 5 red
balls* and 5 green balls*, Box 2 has 7 red balls* and 3
green balls* and Box 3 contains 6 red balls* and 4 green
balls*. The respective probabilities of choosing a box are
1/4, 1/6, 1/8. What is the probability that the ball chosen is
green?
G:= the ball chosen is green.
B1:= Box 1 is selected
Then P(G|B1) = 5/10, P(G|B2) = 3/10 and P(G|B3) = 4/10.
Therefore, using the law of total probability we have
P(G) = P(G|B1) *P(B1) + P(G|B2) P(B2) +P(G|B3) P(B3)
=5/10*1/4+3/10*1/6+4/10*1/8
= 9/40

XP
Bayes’ Theorem
• Let S be a sample space. If A1, A2, A3 ... An are mutually exclusive
and exhaustive events such that P(Ai) ≠ 0 for all i. Then for any
event A which is a subset of
We have

XP
Example
• Let A and B be two events with P(A) = 3/8 ,P(B)=5/8 and P(A
UB)=3/4.Find P(A|B) and P(B|A)
P(A ∩ B) = P(A) +P(B) – P(AUB)
P(A/B) = P(A ∩ B)/P(B)
P(B/A) = P(B) P(A/B) / P(A)

XP
Example
• Three machines A, B and C produce respectively 40%, 10% and
50% of the items in a factory. The % of defective items produced
by the machine is respectively 2%, 3% and 4%. An item from the
factory is selected at random.
– Find the probability that the item is defective.
– If the item is defective, find the probability that the item was
produced by machine C.
• Of all graduate students in university 70% are women and 30%
are men. Suppose that 20% and 25% of the female and male
population respectively smokes cigarettes. What is the probability
that a randomly selected graduate is a) a women who smokes? b)
A man who smokes? c) A smoker?

XP
Three machines A, B and C produce respectively 40%, 10% and 50% of
the items in a factory. The % of defective items produced by the
machine is respectively 2%, 3% and 4%. An item from the factory is
selected at random.
- Find the probability that the item is defective.
- If the item is defective, find the probability that the item was
produced by machine A,B,C.
Example 51

XP
Of all graduate students in university 70% are women and 30% are men.
Suppose that 20% and 25% of the female and male population
respectively smokes cigarettes. What is the probability that a randomly
selected graduate is a) a women who smokes? b) A man who smokes? c)
A smoker?
Example 52
Given: P(F) = 0.70, P(M) = 0.30, P(smoke|F) = 0.20 and P(smoke|M) = 0.25.
a) P(F ∩ smoke) = P(smoke|F)P(F) = 0.20 · 0.70 = 0.14
b) P(M ∩ smoke) = P(smoke|M)P(M) = 0.25 · 0.30 = 0.075
c) P(smoke) = P(smoke|M)P(M) + P(smoke|F)P(F) = 0.14 + 0.075 = 0.215

XP
Example
• An export agency exports tennis balls, which are supplied by 3
manufacturers A, B and C. The balls manufactured by them
contain 3%, 4% and 1% defective balls respectively of the agencies
total export. 50% of balls are manufactured by A, 30% by B, 20%
by C. To test the quality of the balls, one ball is selected at random
and inspected. Find
– P (ball is manufactured by A and is defective)
– P (ball is manufactured by B and is defective)
– P (ball is manufactured by C and is defective)
– P (ball is defective)
– P (ball manufactured by A given that it is defective)
• There are three boxes. Box I contains 1 white, 2 red and 3 black
balls. Box II contains 2 white, 3 red and 1 black balls. Box III
contains 3 white, 1 red and 2 black balls. A box is chosen at
random. If the balls drawn are first red and second white, what is
the probability that they come from Box II?

XP
An export agency exports tennis balls, which are supplied by 3
manufacturers A, B and C. The balls manufactured by them contain 3%,
4% and 1% defective balls respectively of the agencies total export. 50%
of balls are manufactured by A, 30% by B, 20% by C. To test the quality
of the balls, one ball is selected at random and inspected. Find
a) P (ball is manufactured by A and is defective)
b) P (ball is manufactured by B and is defective)
c) P (ball is manufactured by C and is defective)
d) P (ball is defective)
e) P (ball manufactured by A given that it is defective)
Example 53
Given: P(A) = 0.50, P(B) = 0.30, P(C) = 0.20 and
P(D|A) = 0.03, P(D|B) = 0.04 P(D|C) = 0.01.
D is the Event where Ball is defective
a) P(A ∩ D) = P(A)P(D/A) = 0.5 *0.03 = 0.015
b) P(B ∩ D) = P(B)P(D/B) = 0.3 *0.04 = 0.012
c) P(C ∩ D) = P(C)P(D/C) / P(D) = 0.2*0.01 = 0.002
d) P(D) =P(D|A)*P(A)+ P(D|B)*P(B)+P(D|C)*P(C) =
0.03*0.5+0.04*0.3+0.01*0.2
=0.015+0.012+0.002 = 0.029

XP
There are three boxes. Box I contains 1 white, 2 red and 3 black balls.
Box II contains 2 white, 3 red and 1 black balls. Box III contains 3 white, 1
red and 2 black balls. A box is chosen at random. If the balls drawn are
first red and second white, what is the probability that they come from
Box II?
Example 54

XP
In the year 2005 there were three candidates for the position of principal Mr.
Chatterjee, Mr. Iyangar and Mr. Wagh. Their chances of getting the appointment are in
the proportion 4:2:3 respectively. The probability that Mr. Chatterjee is selected would
introduce computer education in the college is 0.3. The probability of Mr. Iyangar and
Mr. Wagh doing the same are respectively 0.5 and 0.8. What is the probability that
there was computer education in the college in 2006?
Example 56

XP
A binary communication channel carries data as one of two types of
signals denoted by 0 and 1. Owing to noise, a transmitted 0 is
sometimes received as 1 and a transmitted 1 is sometimes received as
0. For a given channel, assume a probability of 0.94 that a transmitted 0
is correctly received as a 0 and a probability of 0.91 that a transmitted 1
is received as a 1. Further assume a probability of 0.45 of transferring a
0. If a signal is sent, determine:
Probability that 1 is received.
Probability that 0 is received.
Probability that 1 was transmitted, given that 1 was received.
Probability that 0 was transmitted, given that 0 was received.
Probability of an error.
Example 57

XP
Consider a trinary communication channel whose channel diagram is
shown below. For i = 1, 2, 3, let Ti denote the event “Digit i is
transmitted” and let Ri denote the event “ Digit i is received’. Assume
that a 3 is transmitted three times more frequently than a 1, and a 2 is
sent twice as often as 1. If a 1 has been received , what is the expression
for the probability that a 1 was sent? Derive an expression for the
probability of a transmission error.
Do it Yourself

XP
• A and B are two weak students of statistics and their chances of
solving a problem in statistics correctly are 1/6 and 1/8 resp. If
the probability of their making a common error is 1/525 and they
obtain the same answer, find the probability that their answer is
correct.
• A given lot of IC chips contains 2% defective chips. Each chip is
tested before delivery. The tester itself is not totally reliable so
that:
– P (‘Tester says chip is good”| “ The chip is actually good”) =
0.95
– P (‘Tester says chip is defective”| “ The chip is actually
defective”) = 0.94
– If a tested device is indicated to be defective what is the
probability that it is actually defective?
Do it Yourself

XP
As shown in the fig, The given data is :
P(R1/T1) = 1- α , P(R2/T1) = α/2, P(R3/T1) = α/2
P(R2/T2) = 1- β , P(R3/T2) = β/2, P(R1/T2) = β/2
P(R3/T3) = 1-λ , P(R2/T1) = λ/2, P(R3/T1) = λ/2
Also Given That ,
3 is transmitted three times more frequently than a 1
3P(T1) = P(T3)
2 is sent twice as often as 1
2P(T1) = P(T2)
And p(T1) + P(T2)+P(T3) =1
P(T1)+2P(T1)+3P(T1)=1 so, P(T1) =1/6, P(T2) = 1/3 and P (T3) =1/2
P(R1) = P(R1/T1) *P(T1)+ P(R1/T2) *P(T2)+ P(R1/T3) *P(T3)
=(1- α)*1/6+(β/2)*1/3 +(λ/2)*1/2
a) what is the expression for the probability that a 1 was sent
P(T1/R1) = ?
P(T1/R1) = P(T1) * P(R1/T1) /P(R1)=(1/6)*(1- α)/ ((1- α)*1/6+(β/2)*1/3 +(λ/2b))*1/2)
= α(1- α)/ α(1- α)+2 β+3 λ
b) Derive an expression for the probability of a transmission error
P(Error) = 1-P(Success) = 1- P(R1/T1)T1-P(R2/T2)T2-P(R3/T3)*P(T3)
= 1- (1- α)*1/6- (1- β)1/3- (1-λ)1/2 = (α/6+β/3+λ/2)

XP
A given lot of IC chips contains 2% defective chips. Each chip is tested before delivery.
The tester itself is not totally reliable so that:
P (‘Tester says chip is good”| “ The chip is actually good”) = 0.95
P (‘Tester says chip is defective”| “ The chip is actually defective”) = 0.94
If a tested device is indicated to be defective what is the probability that it is
actually defective?
Example 60
TG = Tester says Chip is Good
AG= Actually good
TD=Test er says chip is defective
AD = Actually defective
Given Data Are:
P(TG/AG)=0.95 -> P(TD/AG)=0.05
P(TD/AD)=0.94
P(AD) = 0.02 -- P(AG)=0.98
P(AD/TD) =??
P(AD/TD) = P(AD) * P(TD/AD) / P(TD/AD)*P(AD) +P(TD/AG)*P(AG)
= 0.02 *0.94 / (0.94*0.02+0.05*0.98)
= 0.2772861

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