Probability+distribution

Basic Probability Concepts
 Probability – the chance that an uncertain event
will occur (always between 0 and 1)
 Impossible Event – an event that has no
chance of occurring (probability = 0)
 Certain Event – an event that is sure to occur
(probability = 1)

Assessing Probability
There are three approaches to assessing the
probability of an uncertain event:
1. a priori -- based on prior knowledge of the process
2. empirical probability
3. subjective probability
outcomeselementaryofnumbertotal
occurcaneventthewaysofnumber
T
X
==
based on a combination of an individual’s past experience,
personal opinion, and analysis of a particular situation
occurcaneventthewaysofnumber
=
Assuming
all
outcomes
are equally
likely
probability of occurrence
probability of occurrence

Example of a priori probability
Find the probability of selecting a face card (Jack,
Queen, or King) from a standard deck of 52 cards.
cardsofnumbertotal
cardsfaceofnumber
CardFaceofyProbabilit ==
T
X
13
3
cardstotal52
cardsface12
==
T
X

Example of empirical probability
Taking Stats Not Taking
Stats
Total
Male 84 145 229
Female 76 134 210
Total 160 279 439
Find the probability of selecting a male taking statistics
from the population described in the following table:
191.0
439
84
peopleofnumbertotal
statstakingmalesofnumber
===Probability of male taking stats

Events
Each possible outcome of a variable is an event.
 Simple event
 An event described by a single characteristic
 e.g., A red card from a deck of cards
 Joint event
 An event described by two or more characteristics
 e.g., An ace that is also red from a deck of cards
 Complement of an event A (denoted A’)
 All events that are not part of event A
 e.g., All cards that are not diamonds

Sample Space
The Sample Space is the collection of all
possible events
e.g. All 6 faces of a die:
e.g. All 52 cards of a bridge deck:

Visualizing Events
 Contingency Tables
 Decision Trees
Red 2 24 26
Black 2 24 26
Total 4 48 52
Ace Not Ace Total
Full Deck
of 52 Cards
Red Card
Black Card
Not an Ace
Ace
Ace
Not an Ace
Sample
Space
Sample
Space2
24
2
24

Visualizing Events
 Venn Diagrams
 Let A = aces
 Let B = red cards
A
B
A ∩ B = ace and red
A U B = ace or red

Definitions
Simple vs. Joint Probability
 Simple Probability refers to the probability of a
simple event.
 ex. P(King)
 ex. P(Spade)
 Joint Probability refers to the probability of an
occurrence of two or more events (joint event).
 ex. P(King and Spade)

Mutually Exclusive Events
 Mutually exclusive events
 Events that cannot occur simultaneously
Example: Drawing one card from a deck of cards
A = queen of diamonds; B = queen of clubs
 Events A and B are mutually exclusive

* Diamonds & Hearts are red in color, while clubs & spades are black
Collectively Exhaustive Events
 Collectively exhaustive events
 One of the events must occur
 The set of events covers the entire sample space
example:
A = aces; B = black cards;
C = diamonds*; D = hearts*
 Events A, B, C and D are collectively exhaustive
(but not mutually exclusive – an ace may also be
a heart)
 Events B, C and D are collectively exhaustive and
also mutually exclusive

Computing Joint and
Marginal Probabilities
 The probability of a joint event, A and B:
 Computing a marginal (or simple) probability:
 Where B1, B2, …, Bk are k mutually exclusive and collectively
exhaustive events
BandAsatisfyingoutcomesofnumber
)BandA(P =
)BdanP(A)BandP(A)BandP(AP(A) k21 +++= 

Joint Probability Example
P(Red and Ace)
Black
Color
Type Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
52
2
cardsofnumbertotal
aceandredarethatcardsofnumber
==

Marginal Probability Example
P(Ace)
Black
Color
Type Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
52
4
52
2
52
2
)BlackandAce(P)dReandAce(P =+=+=

P(A1 and B2) P(A1)
TotalEvent
Marginal & Joint Probabilities In A
Contingency Table
P(A2 and B1)
P(A1 and B1)
Event
Total 1
Joint Probabilities Marginal (Simple) Probabilities
A1
A2
B1 B2
P(B1) P(B2)
P(A2 and B2) P(A2)

Probability Summary So Far
 Probability is the numerical measure
of the likelihood that an event will
occur
 The probability of any event must be
between 0 and 1, inclusively
 The sum of the probabilities of all
mutually exclusive and collectively
exhaustive events is 1
Certain
Impossible
0.5
1
0
0 ≤ P(A) ≤ 1 For any event A
1P(C)P(B)P(A) =++
If A, B, and C are mutually exclusive and
collectively exhaustive

General Addition Rule
P(A or B) = P(A) + P(B) - P(A and B)
General Addition Rule:
If A and B are mutually exclusive, then
P(A and B) = 0, so the rule can be simplified:
P(A or B) = P(A) + P(B)
For mutually exclusive events A and B

General Addition Rule Example
P(Red or Ace) = P(Red) +P(Ace) - P(Red and Ace)
= 26/52 + 4/52 - 2/52 = 28/52
Don’t count
the two red
aces twice!
Black
Color
Type Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52

Quiz Time!
 Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random.
What is the probability that the ticket drawn has a number which is a multiple
of 3 or 5?
 A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at
random. What is the probability that none of the balls drawn is blue?
 What is the probability of getting a sum 9 from two throws of a dice?
 Two dice are thrown simultaneously. What is the probability of getting two
numbers whose product is even?
 From a pack of 52 cards, two cards are drawn together at random. What is
the probability of both the cards being kings?
 A card is drawn from a pack of 52 cards. What is the probability of getting a
queen of club or a king of heart?
 You are supposed to attend 3 job interviews. For the first company, there are
12 candidates for one position; for the second, there are 15 candidates for two
positions, while for the third there are 20 candidates for 4 positions. What are
the chances that you will get at least one offer?

Computing Conditional
Probabilities
 A conditional probability is the probability of one
event, given that another event has occurred:
P(B)
B)andP(A
B)|P(A =
P(A)
B)andP(A
A)|P(B =
Where P(A and B) = joint probability of A and B
P(A) = marginal or simple probability of A
P(B) = marginal or simple probability of B
The conditional
probability of A given
that B has occurred
The conditional
probability of B given
that A has occurred

 What is the probability that a car has a CD
player, given that it has AC ?
i.e., we want to find P(CD | AC)
Conditional Probability Example
 Of the cars on a used car lot, 70% have air
conditioning (AC) and 40% have a CD player
(CD). 20% of the cars have both.

No CDCD Total
AC 0.2 0.5 0.7
No AC 0.2 0.1 0.3
Total 0.4 0.6 1.0
 Of the cars on a used car lot, 70% have air conditioning
(AC) and 40% have a CD player (CD).
20% of the cars have both.
0.2857
0.7
0.2
P(AC)
AC)andP(CD
AC)|P(CD ===
(continued)

No CDCD Total
AC 0.2 0.5 0.7
No AC 0.2 0.1 0.3
Total 0.4 0.6 1.0
 Given AC, we only consider the top row (70% of the cars). Of these,
20% have a CD player. 20% of 70% is about 28.57%.
0.2857
0.7
0.2
P(AC)
AC)andP(CD
AC)|P(CD ===
(continued)

Using Decision Trees
Has AC
Does nothave AC
Has CD
Does nothave CD
Has CD
Does nothave CD
P(AC)= 0.7
P(AC’)= 0.3
P(AC and CD) = 0.2
P(AC and CD’) = 0.5
P(AC’ and CD’) = 0.1
P(AC’ and CD) = 0.2
7.
5.
3.
2.
3.
1.
All
Cars
7.
2.
Given AC or
no AC:
Conditional
Probabilities

Using Decision Trees
Has CD
Does nothave CD
Has AC
Does nothave AC
Has AC
Does nothave AC
P(CD)= 0.4
P(CD’)= 0.6
P(CD and AC) = 0.2
P(CD and AC’) = 0.2
P(CD’ and AC’) = 0.1
P(CD’ and AC) = 0.5
4.
2.
6.
5.
6.
1.
All
Cars
4.
2.
Given CD or
no CD:
(continued)
Conditional
Probabilities

 In AGBS Bangalore, 65% of MBA students took
up Marketing as their specialization, while 50%
took up Finance. It is also known that 20% of the
students opted for Marketing but did not take
Finance.
 What is the probability that a student takes
Marketing, given that he takes Finance?

 A machine produces parts that are either good
(90%), slightly defective (2%), or obviously
defective (8%). Produced parts get passed
through an automatic inspection machine, which
is able to detect any part that is obviously
defective and discard it.
 What is the probability that a ‘good’ part makes it
through the inspection machine and gets
shipped?

 Your neighbor has 2 children. You learn that he has a
son, Joe.
 What is the probability that Joe’s sibling is a brother?

Joe again!
 Your neighbor has 2 children. He picks one of them at random and
comes by your house; he brings a boy named Joe (his son). What
is the probability that Joe’s sibling is a brother?
 This is not the same as the previous problem. How?!
 P(Second also a boy given that first one turned up to be a boy
when randomly selected) = P(1st
one boy & 2nd
one boy)/P(first
one is a boy on random selection)
= (1/2*1/2)/(1/2) = 1/2

Independence
 Two events are independent if and only
if:
 Events A and B are independent when the probability
of one event is not affected by the fact that the other
event has occurred
P(A)B)|P(A =

Multiplication Rules
 Multiplication rule for two events A and B:
P(B)B)|P(AB)andP(A =
P(A)B)|P(A =Note: If A and B are independent, then
and the multiplication rule simplifies to
P(B)P(A)B)andP(A =

Multiplication Rules [contd…]

Example 1
 Suppose that five good fuses and two defective ones have been
mixed up. To find the defective fuses, we test them one-by-one, at
random and without replacement. What is the probability that we
are lucky and find both of the defective fuses in the first two
tests?

Example 2
 If six cards are selected at random (without replacement) from a
standard deck of 52 cards, what is the probability there will be no
pairs? (two cards of the same denomination)?

Marginal Probability
 Marginal probability for event A:
 Where B1, B2, …, Bk are k mutually exclusive and
collectively exhaustive events
)P(B)B|P(A)P(B)B|P(A)P(B)B|P(AP(A) kk2211 +++= 

Solve this!
 The probability that each relay closes in the circuit shown below is
p. Assuming that each relay functions independently of the others,
find the probability that current can flow from L to R.

Now solve this!
 A high school conducts random drug tests on its students. Of the
student body, it is known that 8% use marijuana regularly; 17%
use it occasionally; and 75% never use it. The testing regime is
not perfect: regular marijuana users falsely test negative 5% of
the time; occasional users falsely test negative 13% of the time;
and non-users falsely test positive 11% of the time. What
percentage of the student body will test positive for marijuana use?

Bayes’ Theorem
 Bayes’ Theorem is used to revise previously
calculated probabilities based on new
information.
 Developed by Thomas Bayes in the 18th
Century.
 It is an extension of conditional probability.

Bayes’ Theorem Example 1
 A drilling company has estimated a 40%
chance of striking oil for their new well.
 A detailed test has been scheduled for more
information. Historically, 60% of successful
wells have had detailed tests, and 20% of
unsuccessful wells have had detailed tests.
 Given that this well has been scheduled for a
detailed test, what is the probability that the
well will be successful?

 Let S = successful well
U = unsuccessful well
 P(S) = 0.4 , P(U) = 0.6 (prior probabilities)
 Define the detailed test event as D
 Conditional probabilities:
P(D|S) = 0.6 P(D|U) = 0.2
 Goal is to find P(S|D)
(continued)

0.667
0.120.24
0.24
(0.2)(0.6)(0.6)(0.4)
(0.6)(0.4)
U)P(U)|P(DS)P(S)|P(D
S)P(S)|P(D
D)|P(S
=
+
=
+
=
+
=
(continued)
Apply Bayes’ Theorem:
So the revised probability of success, given that this well
has been scheduled for a detailed test, is 0.667

 Urn 1 contains 5 white balls and 7 black balls.
Urn 2 contains 3 whites and 12 black. A fair
coin is flipped; if it is Heads, a ball is drawn
from Urn 1, and if it is Tails, a ball is drawn
from Urn 2. Suppose that this experiment is
done and you learn that a white ball was
selected.
 What is the probability that this ball was in fact
taken from Urn 2? (i.e., that the coin flip was
Tails)

 Bob can decide to go to work by one of three modes of
transportation, car, bus, or commuter train. Because of high
traffic, if he decides to go by car, there is a 50% chance he will be
late. If he goes by bus, which has special reserved lanes but is
sometimes overcrowded, the probability of being late is only 20%.
The commuter train is almost never late, with a probability of only
1%, but is more expensive than the bus.
(a) Suppose that Bob is late one day, and his boss wishes to estimate the
probability that he drove to work that day by car. Since he does not know
which mode of transportation Bob usually uses, he gives a prior probability
of 1/3 to each of the three possibilities. What is the boss’ estimate of the
probability that Bob drove to work?

 Bob can decide to go to work by one of three modes of
transportation, car, bus, or commuter train. Because of high
traffic, if he decides to go by car, there is a 50% chance he will be
late. If he goes by bus, which has special reserved lanes but is
sometimes overcrowded, the probability of being late is only 20%.
The commuter train is almost never late, with a probability of only
1%, but is more expensive than the bus.
(b) Suppose that a coworker of Bob’s knows that he almost always takes the
commuter train to work, never takes the bus, but sometimes, 10% of the
time, takes the car. What is the coworkers probability that Bob drove to
work that day, given that he was late?

 A company has two plants to manufacture motorcycles. 70%
motor cycles are manufactured at the first plant, while 30% are
manufactured at the second plant. At the first plant, 80%
motorcycles are rated of the standard quality while at the second
plant, 90% are rated of standard quality. A motorcycle, randomly
picked up, is found to be of standard quality.
 Find the probability that it has come out from the second plant..

 A manufacturing firm produces steel pipes in three plants A,B and
C with daily production of 500,1000 and 2000 units respectively.
The fractions of defective steel pipes output produced by the
plants A,B and C are respectively .005, .008 and .010.
 If a pipe is selected from a day's total production and found to be
defective, find out the probability that it came from the first plant.
 The chance that the defective pipe came from a particular plant is the
highest for which plant?

 At a certain university, 4% of men are over 6 feet tall and 1% of
women are over 6 feet tall. The total student population is divided
in the ratio 3:2 in favour of women.
 If a student is selected at random from among all those over six feet tall,
what is the probability that the student is a woman?

 Suppose that we have two identical boxes: box 1 and box 2. Box
1 contains 5 red balls and 3 blue balls. Box 2 contains 2 red balls
and 4 blue balls. A box is selected at random and exactly one ball
is drawn from the box.
 Given that the selected ball is blue, what’s the probability that it
came from box 2?

 An insurance company divides its policy holders into three
categories: low risk, moderate risk, and high risk. The low-risk
policy holders account for 60% of the total number of people
insured by the company. The moderate-risk policy holders
account for 30%, and the high-risk policy holders account for
10%. The probabilities that a low-risk, moderate-risk, and high-
risk policy holder will file a claim within a given year are
respectively .01, .10, and .50
 Given that a policy holder files a claim this year, what is the
probability that the person is a high-risk policy holder??

Counting Rules
 Rules for counting the number of possible
outcomes
 Counting Rule 1:
 If any one of k different mutually exclusive and
collectively exhaustive events can occur on each of
n trials, the number of possible outcomes is equal to
 Example

If you roll a fair die 3 times then there are 63
= 216 possible
outcomes
kn

Counting Rules
 If there are k1 events on the first trial, k2 events on the
second trial, … and kn events on the nth
trial, the
number of possible outcomes is
 Example:

You want to go to a park, eat at a restaurant, and see a
movie. There are 3 parks, 4 restaurants, and 6 movie
choices. How many different possible combinations are
there?

Answer: (3)(4)(6) = 72 different possibilities
(k1)(k2)…
(kn)
(continued)

Counting Rules
 The number of ways that n items can be arranged in
order is
 Example:

You have five books to put on a bookshelf. How many
different ways can these books be placed on the shelf?

Answer: 5! = (5)(4)(3)(2)(1) = 120 different possibilities
n! = (n)(n – 1)…
(1)
(continued)

Counting Rules
 Permutations: The number of ways of arranging X
objects selected from n objects in order is
 Example:

You have five books and are going to put three on a
bookshelf. How many different ways can the books be
ordered on the bookshelf?

Answer: different possibilities
(continued)
X)!(n
n!
Pxn
−
=
60
2
120
3)!(5
5!
X)!(n
n!
Pxn ==
−
=
−
=

Counting Rules
 Combinations: The number of ways of selecting X
objects from n objects, irrespective of order, is
 Example:

You have five books and are going to select three are to
read. How many different combinations are there, ignoring
the order in which they are selected?

Answer: different possibilities
(continued)
X)!(nX!
n!
Cxn
−
=
10
(6)(2)
120
3)!(53!
5!
X)!(nX!
n!
Cxn ==
−
=
−
=

Probability+distribution

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