One Fourth Labs
We deliver courseware in AI and related areas
Recap: List of Topics
Descriptive Statistics
Probability Theory
Inferential Statistics
Different types of data
Different types of plots
Measures of centrality and spread
Counting, Sample spaces, events
Discrete and continuous RVs
Bernoulli, Uniform, Normal dist.
Sampling strategies
Interval Estimators
Hypothesis testing (z-test, t-test)
ANOVA, Chi-square test
Linear Regression
Learning Objectives
What are sets and some of their properties
What are experiments, sample spaces, outcomes and events?
What are the axioms of probability?
What are some simple ways of defining a probability function?
What are some important theorems: multiplication rule, total probability theorem and Bayes' theorem?
What are independent events?
The element of chance (Nothing in life is certain)
Randomness everywhere!
What is the chance that he would get infected if he went to the market?
No definite answer!
Why?
Due to the random nature of the world around us
Randomness everywhere!
What is the mode of transport ?
Is private car always safer than public transport?
How good is his immune system?
Does he have co-morbidities?
How many infections in the neighbourhood?
Randomness everywhere!
How many infections in the neighbourhood?
Randomness everywhere!
How many infections in the neighbourhood?
The study of this chance is the subject matter of Probability Theory!
Set theory
Experiments, sample spaces, events
Axioms of Probability
Random Variables
Distributions
Expectation
A brief overview of Set Theory
Set: A collection of elements
S={a,e,i,o,u}
S = \{a,e,i,o,u\}
E={0,2,4,…,96,98,100}
E = \{0,2,4,\dots,96,98,100\}
E={x:0≤x≤100,x%2=0}
E = \{x: 0 \leq x \leq 100, x\%2 = 0\}
Compact notation: Useful for large sets
x∈S
x \in S
mean x belongs to the set S
2∈E, 3∈/E
2 \in E,~~3 \notin E
Subsets and equal sets
I:set of all integers
\mathbb{I}: set~of~all~integers
S={x:x∈I,x<0}
S = \{x: x \in \mathbb{I}, x < 0 \}
Every element of in contained in
S
S
I
\mathbb{I}
S⊂I
S \subset \mathbb{I}
A=B iff A⊂B and B⊂A
A = B~~iff~~A \subset B~and~B \subset A
subset
equal sets
Universal Set
A:set of all aces
A: set~of~all~aces
Ω=set of all 52 cards
\Omega = set~of~all~52~cards
Every set of interest is a subset of the universal set
H:set of all hearts
H: set~of~all~hearts
B:set of all black cards
B: set~of~all~black~cards
F:set of all face cards
F: set~of~all~face~cards
A⊂Ω
A \subset \Omega
H⊂Ω
H \subset \Omega
B⊂Ω
B \subset \Omega
F⊂Ω
F \subset \Omega
Empty Set
ϕ={}
\phi = \{\}
Set with no elements (null set)
Set Operations
Ac={x:x∈Ω and x∈/A}
A^\mathsf{c} = \{x: x \in \Omega~and~x\notin A\}
Complement
Union (2 sets)
A∪B={x:x∈A or x∈B}
A \cup B = \{x: x \in A~or~x\in B\}
(black, gray and white in the image)
Intersection (2 sets)
A∩B={x:x∈A and x∈B}
A \cap B = \{x: x \in A~and~x\in B\}
(gray area in the image)
Set Operations (n sets)
Union (n sets)
x∈A1∩A2∩A3⋯∩An iff x∈Ai∀i
x \in A_1 \cap A_2 \cap A_3 \cdots \cap A_n ~iff~x \in A_i \forall i
Intersection (n sets)
x∈A1∪A2∪A3⋯∪An iff x∈Aifor some i
x \in A_1 \cup A_2 \cup A_3 \cdots \cup A_n ~iff~x \in A_i for~some~i
Ω
\Omega
Properties of Set operations
Commutativity
A∪B=B∪A
A \cup B = B \cup A
A∩B=B∩A
A \cap B = B \cap A
Associativity
A∪(B∪C)=(A∪B)∪C
A \cup (B \cup C) = (A \cup B) \cup C
A∩(B∩C)=(A∩B)∩C
A \cap (B \cap C) = (A \cap B) \cap C
Ω
\Omega
Properties of Set operations
Distributive Laws
A∩(B∪C)=(A∩B)∪(A∩C)
A \cap (B \cup C) = (A \cap B) \cup (A \cap C)
Proof
A∪(B∩C)=(A∪B)∩(A∪C)
A \cup (B \cap C) = (A \cup B) \cap (A \cup C)
x∈A∩(B∪C)
x \in A\cap (B \cup C)
⟹x∈A and x∈(B∪C)
\implies x \in A~and~x \in (B \cup C)
⟹x∈A and B or x∈A and C
\implies x \in A~and~B~or~x \in A~and~C
Ω
\Omega
Properties of Set operations
DeMorgan's Laws
(A∪B)c=Ac∩Bc
(A \cup B)^\mathsf{c} = A^\mathsf{c} \cap B^\mathsf{c}
Proof
⟹x∈Ac∪Bc
\implies x \in A^\mathsf{c} \cup B^\mathsf{c}
(A∩B)c=Ac∪Bc
(A \cap B)^\mathsf{c} = A^\mathsf{c} \cup B^\mathsf{c}
x∈(A∩B)c
x \in (A \cap B)^\mathsf{c}
⟹x∈/A∩B
\implies x \notin A \cap B
⟹x∈/A or x∈/B
\implies x\notin A~or~x\notin B
⟹x∈Ac or x∈Bc
\implies x \in A^\mathsf{c} ~or~x \in B^\mathsf{c}
Ω
\Omega
A
A
B
B
Countable v/s Uncountable Infinite Sets
: Set of all real numbers has infinite elements
: Set of all integers has infinite elements
I
\mathbb{I}
R
\mathbb{R}
(uncountable)
(countable)
An infinite set is said to be countable if there is a 1-1 correspondence between the elements of this set and the set of positive integers
Countable Infinite Sets
I={−∞,…,−3,−2,−1,0,1,2,3,…,∞}
\mathbb{I} = \{-\infty, \dots, -3, -2, -1, 0, 1, 2, 3, \dots, \infty\}
={0,1,−1,2,−2,3,−3…}
= \{0,1,-1,2,-2,3,-3 \dots \}
: Set of all positive rational numbers
P
\mathbb{P}
P={21,31,32,41,42,43,…}
\mathbb{P} = \{\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \dots\}
1 2 3 4 5 6 7 ....
1 2 3 4 5 ....
Uncountable Infinite Sets
: Set of all real numbers
R
\mathbb{R}
Q=[0,1]
\mathbb{Q} = [0,1]
Experiments and Sample Spaces
Experiment: Bowling a bowl
Outcome: {0, 1, 2, 3, 4, 5, 6} runs
Experiment: Going to the mall
Outcome: {infected, not_infected}
Experiment: Blood Test
Outcome: {positive, negative}
Experiment: Writing an exam
Outcome: {A, B, C, D, E, F}
The outcome in every trial is uncertain but the set of outcomes is certain
An experiment or trial is any procedurethat can be repeated infinite times and has a well-defined set of outcomes
The set of all possible outcomes of an experiment is called the sample space. The elements in a sample space are mutually exclusive and collectively exhaustive
Experiments involving coin tosses
Ω={H,T}
\Omega = \{H, T\}
Ω={HH,
\Omega = \{HH,
HT,
HT,
TH,
TH,
TT}
TT\}
Ω={HHH,
\Omega = \{HHH,
HHT,
HHT,
HTH,HTT,
HTH, HTT,
THH,THT,
THH, THT,
TTH,TTT}
TTH, TTT\}
∣Ω∣=2
|\Omega| = 2
∣Ω∣=4
|\Omega| = 4
∣Ω∣=8
|\Omega| = 8
…n coins
\dots n~coins
∣Ω∣=2n
|\Omega| = 2^n
Experiments involving fair dice
Ω={1,2,3,4,5,6}
\Omega = \{1, 2, 3, 4, 5, 6\}
∣Ω∣=6
|\Omega| = 6
∣Ω∣=36
|\Omega| = 36
…n dice
\dots n~dice
∣Ω∣=6n
|\Omega| = 6^n
Ω={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
\Omega = \{ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), \newline (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),\newline(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),\newline (4,1), (4,2), (4,3), (4,4), (4,5), (4,6),\newline (5,1), (5,2), (5,3), (5,4), (5,5), (5,6),\newline (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \}
Experiments involving cards
∣Ω∣=52
|\Omega| = 52
∣Ω∣=523
|\Omega| = 52^3
∣Ω∣=522
|\Omega| = 52^2
∣Ω∣=524
|\Omega| = 52^4
More the number of outcomes, less the probability of any single outcome (assuming all outcomes are equally likely)
Experiments:continuous outcomes
Ω={(x,y) s.t. 0≤x,y≤1}
\Omega = \{(x,y)~s.t.~0\leq x,y\leq 1\}
Events of an experiment
Ω={HH,HT,TH,TT}
\Omega = \{HH, HT, TH, TT\}
An event is a set of outcomes of an experiment. This set is a subset of the sample space
A={HH,HT}
A = \{HH, HT\}
(the event that the first toss results in a head)
B={TT}
B = \{TT\}
(the event that both the tosses result in tails)
(the event that there are exactly 2 aces)
∣C∣=(24)∗(148)=288
|C| = {4 \choose 2} * {48 \choose 1} = 288
We say that event A has occurred if the outcome of the experiment lies in the set A
A={HH,HT}
A = \{HH, HT\}
(the event that the first toss results in a head)
B={TT}
B = \{TT\}
(the event that both the tosses result in tails)
(the event that there are exactly 2 aces)
∣C∣=(24)∗(148)=288
|C| = {4 \choose 2} * {48 \choose 1} = 288
Ω={HH,HT,TH,TT}
\Omega = \{HH, HT, TH, TT\}
Union of events
A={(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)}
A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)\}
B={(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)}
B = \{(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\}
C=A∪B
C = A\cup B
(the event that the first die shows a 2)
(the event that the second die shows a 4)
={(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), (1,4),(2,4),(3,4),(4,4),(5,4),(6,5)}
= \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),\\~~~~~~~~ (1,4), (2,4), (3,4), (4,4), (5,4), (6,5)\}
(the event that the first die shows a 2 or the second die shows a 4)
Intersection of events
A={(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)}
A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)\}
B={(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)}
B = \{(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\}
D=A∩B
D = A \cap B
(the event that the first die shows a 2)
(the event that the second die shows a 4)
(the event that the first die shows a 2 and the second die shows a 4)
={(2,4)}
= \{(2,4)\}
Complement of events
A={(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)}
A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)\}
B={(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)}
B = \{(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\}
E=Ac
E = A^\mathsf{c}
(the event that the first die shows a 2)
(the event that the second die shows a 4)
(the event that the first die does not show a 2)
Multiple events
A
A
: the hand contains the ace of spade
B
B
: the hand contains the ace of clubs
C
C
: the hand contains the ace of hearts
A∪B∪C
A \cup B \cup C
A∩B∩C
A \cap B \cap C
Ω
\Omega
A
A
B
B
C
C
Disjoint events
A
A
: the event that the first die shows a 1
B
B
: the event that the first die shows a 2
A and Ac
A~and~A^\mathsf{c}
Two events A and B are said to be disjoint if they cannot occur simultaneously, i.e.,
A∩Ac=ϕ
A\cap A^\mathsf{c}= \phi
A∩B=ϕ
A\cap B= \phi
A∩B=ϕ
A\cap B= \phi
A∪Ac=Ω
A\cup A^\mathsf{c}= \Omega
A∪B=Ω
A\cup B \neq \Omega
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), \newline (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),\newline(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),\newline (4,1), (4,2), (4,3), (4,4), (4,5), (4,6),\newline (5,1), (5,2), (5,3), (5,4), (5,5), (5,6),\newline (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Ω
\Omega
A
A
B
B
Disjoint events
A={HH}
A = \{HH\}
B={TT}
B=\{TT\}
The events are said to be mutually disjoint or pairwise disjoint if
A∩B=ϕ
A\cap B= \phi
Ai∩Aj=ϕ ∀ i,j s.t. i=j
A_i\cap A_j= \phi~\forall~i,j~s.t.~i\neq j
B∩C=ϕ
B\cap C = \phi
A1,A2,…,An
A_1, A_2, \dots, A_n
C={HT,TH}
C=\{HT,TH\}
A∩C=ϕ
A\cap C = \phi
Partition of the sample space
A={HH}
A = \{HH\}
B={TT}
B=\{TT\}
If the events are mutually disjoint and then are said to partition the sample space
A∩B=ϕ
A\cap B= \phi
B∩C=ϕ
B\cap C = \phi
A1,A2,…,An
A_1, A_2, \dots, A_n
C={HT,TH}
C=\{HT,TH\}
A∩C=ϕ
A\cap C = \phi
A1∪A2∪⋯∪An=Ω
A_1 \cup A_2 \cup \dots \cup A_n = \Omega
A1,A2,…,An
A_1, A_2, \dots, A_n
A∪B∪C=Ω
A\cup B\cup C = \Omega
A1
A_1
A5
A_5
A4
A_4
A3
A_3
A2
A_2
A6
A_6
A7
A_7
Axioms of Probability
Axioms of Probability
Recap
Experiments
Sample Space
Events
What is the chance of an event?
Goal: Assign a number to each event such that this number reflects the chance of the experiment resulting in that event
The probability function
What are the conditions that such a probability function must satisfy?
(Axioms of Probability)
P(A)=?
P(A) = ?
Event
Probability function
The axioms of probability
P(A)≥0 ∀A
P(A) \geq 0~\forall A
Axiom 1 (non-negativity)
P(Ω)=1
P(\Omega) = 1
Axiom 2 (normalisation)
Axiom 3 (finite additivity)
If the events are mutually disjoint then
A1,A2,…,An
A_1, A_2, \dots, A_n
=∑i=1nP(Ai)
= \sum_{i=1}^n P(A_i)
P(A1∪A2∪⋯∪An)
P(A_1\cup A_2 \cup \dots \cup A_n)
The axioms of probability
Axiom 3 (finite additivity)
=∑i=1nP(Ai)
= \sum_{i=1}^n P(A_i)
P(A1∪A2∪⋯∪An)
P(A_1\cup A_2 \cup \dots \cup A_n)
Smallest possible event = one outcome
Compute probabilities of larger events from smaller events
A1
A_1
A2
A_2
A3
A_3
A4
A_4
A5
A_5
A6
A_6
The axioms of probability
Given
P(A1),P(A2),P(A3),P(A4),P(A5),P(A6)
P(A_1),P(A_2),P(A_3),P(A_4),P(A_5),P(A_6)
B
B
: the event that the outcome is an odd no.
C
C
: the event that the outcome is
P(B)=P(A1)+P(A3)+P(A5)
P(B) = P(A_1)+P(A_3)+P(A_5)
≥5
\geq 5
P(C)=P(A5)+P(A6)
P(C) = P(A_5)+P(A_6)
D
D
: the event that the outcome is a mult. of 3
P(D)=P(A3)+P(A6)
P(D) = P(A_3)+P(A_6)
we can compute other probabilities
The probability of an event can be computed as the sum of the probabilities of the disjoint outcomes contained in the event
Some properties of probability
Property 1:
P(A)=1−P(Ac)
P(A) = 1 - P(A^\mathsf{c})
A∪Ac=Ω
A \cup A^\mathsf{c} = \Omega
1=P(Ω)=P(A∪Ac)=P(A)+P(Ac)
1 = P(\Omega) = P(A \cup A^\mathsf{c}) = P(A) + P(A^\mathsf{c})
∴P(A)=1−P(Ac)
\therefore P(A) = 1 - P(A^\mathsf{c})
Property 2:
P(A)≤1
P(A) \leq 1
P(A)=1−P(Ac)
P(A) = 1 - P(A^\mathsf{c})
Some properties of probability
Property 3:
P(A∪B)=P(A)+P(B)−P(A∩B)
P(A \cup B) = P(A) + P(B) - P(A \cap B)
Ac
A^\mathsf{c}
P(A∪B)=P(A∪(B∩Ac))
P(A \cup B) = P(A \cup (B \cap A^\mathsf{c}))
=P(A)+P(B∩Ac)
= P(A) + P(B \cap A^\mathsf{c})
P(B)=P((B∩Ac)∪(B∩A))
P(B) = P((B \cap A^\mathsf{c}) \cup (B \cap A))
=P(B∩Ac)+P(B∩A)
= P(B\cap A^\mathsf{c}) + P(B\cap A)
∴P(B∩Ac)=P(B)−P(B∩A)
\therefore P(B\cap A^\mathsf{c}) = P(B) - P(B\cap A)
=P(A)+P(B)−P(B∩A)
= P(A) + P(B) - P(B\cap A)
Some properties of probability
Property 4:
Ω=A1∪A2∪⋯∪An
\Omega = A_1 \cup A_2 \cup \dots \cup A_n
the sum of the probabilities of all outcomes is equal to 1
P(Ω)=P(A1∪A2∪⋯∪An)=∑i=1nP(Ai)
P(\Omega) = P(A_1 \cup A_2 \cup \dots \cup A_n) = \sum_{i=1}^n P(A_i)
∴∑i=1nP(Ai)=1
\therefore \sum_{i=1}^n P(A_i) = 1
\underbrace{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}
Ω
\Omega
Property 5:
P(ϕ)=0
P(\phi) = 0
∴1=P(Ω)=P(Ω∪ϕ)=P(Ω)+P(ϕ)
\therefore 1 = P(\Omega) = P(\Omega \cup \phi) = P(\Omega) + P(\phi)
∴1=1+P(ϕ)⟹P(ϕ)=0
\therefore 1 = 1 + P(\phi) \implies P(\phi) = 0
Examples
Outcomes: {0, 1, 2, 3, 4, 5, 6}
A0,A1,A2,A3,A4,A5,A6
A_0, A_1, A_2, A_3, A_4, A_5, A_6
P(A0)=0.3,P(A1)=0.45,P(A2)=0.12,P(A3)=0.02,P(A4)=0.07,P(A5)=0.01,P(A6)=0.03
P(A_0)=0.3, P(A_1)=0.45, P(A_2)=0.12, P(A_3)=0.02, \\ P(A_4)=0.07, P(A_5)=0.01, P(A_6)=0.03
What is the prob. of scoring even no. of runs?
X=A0∪A2∪A4∪A6
X = A_0 \cup A_2 \cup A_4 \cup A_6
P(X)=P(A0∪A2∪A4∪A6)
P(X) = P(A_0 \cup A_2 \cup A_4 \cup A_6)
=P(A0)+P(A2)+P(A4)+P(A6)
= P(A_0) + P(A_2) + P(A_4) + P(A_6)
Examples
Outcomes: {0, 1, 2, 3, 4, 5, 6}
A0,A1,A2,A3,A4,A5,A6
A_0, A_1, A_2, A_3, A_4, A_5, A_6
P(A0)=0.3,P(A1)=0.45,P(A2)=0.12,P(A3)=0.02,P(A4)=0.07,P(A5)=0.01,P(A6)=0.03
P(A_0)=0.3, P(A_1)=0.45, P(A_2)=0.12, P(A_3)=0.02, \\ P(A_4)=0.07, P(A_5)=0.01, P(A_6)=0.03
What is the prob. of scoring less than 5 runs?
X=A0∪A1∪A2∪A3∪A4
X = A_0 \cup A_1 \cup A_2 \cup A_3 \cup A_4
P(X)=P(A0∪A1∪A2∪A3∪A4)
P(X) = P(A_0 \cup A_1 \cup A_2 \cup A_3 \cup A_4)
=P(A0)+P(A1)+P(A2)+P(A3)+P(A4)
= P(A_0) + P(A_1) + P(A_2) + P(A_3) + P(A_4)
Examples
Outcomes: {0, 1, 2, 3, 4, 5, 6}
A0,A1,A2,A3,A4,A5,A6
A_0, A_1, A_2, A_3, A_4, A_5, A_6
P(A0)=0.3,P(A1)=0.45,P(A2)=0.12,P(A3)=0.02,P(A4)=0.07,P(A5)=0.01,P(A6)=0.03
P(A_0)=0.3, P(A_1)=0.45, P(A_2)=0.12, P(A_3)=0.02, \\ P(A_4)=0.07, P(A_5)=0.01, P(A_6)=0.03
The prob. that the runs will be div. by 2 or 3?
X1={0,2,4,6}=A0∪A2∪A4∪A6
X_1 = \{0, 2, 4, 6\} = A_0 \cup A_2 \cup A_4 \cup A_6
X2={0,3,6}=A0∪A3∪A6
X_2 = \{0, 3, 6\} = A_0 \cup A_3 \cup A_6
X1∩X2={0,6}=A0∪A6
X_1 \cap X_2 = \{0, 6\} = A_0 \cup A_6
P(X1)=0.3+0.12+0.07+0.03=0.52
P(X_1) = 0.3 + 0.12 + 0.07 + 0.03 = 0.52
P(X2)=0.3+0.02+0.03=0.35
P(X_2) = 0.3 + 0.02 + 0.03 = 0.35
P(X1∩X2)=0.3+0.03=0.33
P(X_1 \cap X_2) = 0.3 + 0.03 = 0.33
P(X1∪X2)=P(X1)+P(X2)−P(X1∩X2)
P(X_1 \cup X_2) = P(X_1) + P(X_2) - P(X_1 \cap X_2)
∴P(X1∪X2)=0.52+0.35−0.33=0.54
\therefore P (X_1 \cup X_2) = 0.52 + 0.35 - 0.33 = 0.54
Examples
The ball bearings manufactured by a factory have two types of defects. Suppose the probability of having Type 1 defect is 0.01, having Type 2 defect is 0.02 and having both is 0.0005
P(A1)=0.01,P(A2)=0.03,P(A1∩A2)=0.005
P(A_1)=0.01, P(A_2)=0.03, P(A_1 \cap A_2) = 0.005
P(A1∪A2)=P(A1)+P(A2)−P(A1∩A2)
P(A_1 \cup A_2)= P(A_1) + P(A_2) - P(A_1 \cap A_2)
What is the prob. that the bearing will have type 1 or type 2 defect ?
=0.01+0.02−0.005=0.025
= 0.01 + 0.02 - 0.005 = 0.025
A2
A_2
A1
A_1
Examples
The ball bearings manufactured by a factory have two types of defects. Suppose the probability of having Type 1 defect is 0.01, having Type 2 defect is 0.02 and having both is 0.0005
P(A1)=0.01,P(A2)=0.03,P(A1∩A2)=0.005
P(A_1)=0.01, P(A_2)=0.03, P(A_1 \cap A_2) = 0.005
P(A1∪A2)c=1−P(A1∪A2)
P(A_1 \cup A_2)^\mathsf{c}= 1 - P(A_1 \cup A_2)
What is the prob. that the bearing will have neither type 1 nor type 2 defect ?
=1−0.025=0.975
= 1 - 0.025 = 0.975
A2
A_2
A1
A_1
Designing Probability Functions
Designing Probability Functions
(probability as relative frequency)
Recap
Goal: Assign a number to each event such that this number reflects the chance of the experiment resulting in that event
Required: The probability function must satisfy the axioms of probability
Probability as relative frequency
Karl Pearson tossed a coin 24000 times he observed that the number of heads was 12012
P(H)=2400012012=0.5005
P(H) = \frac{12012}{24000} = 0.5005
We can think of the probability of an event as the fraction of times the event occurs when an experiment is repeated a large number of times
Probability as relative frequency
We can think of the probability of an event as the fraction of times the event occurs when an experiment is repeated a large number of times
P(Ai)=total no. of times the experiment was repeatedno. of times the outcome is in Ai
P(A_i) = \frac{no.~of~times~the~outcome~is~in~A_i}{total~no.~of~times~the~experiment~was~repeated}
But does such a P() satisfy the axioms of probability?
Probability as relative frequency
P(Ai)≥0 ?
P(A_i) \geq 0~?
: ratio of two positive numbers
Does P() satisfy the axioms?
P(Ω)=1 ?
P(\Omega)=1~?
P(Ω)=total no. of times the experiment was repeatedno. of times the outcome is in S=1
P(\Omega) = \frac{no.~of~times~the~outcome~is~in~S}{total~no.~of~times~the~experiment~was~repeated} = 1
Ω
\Omega
P(A1∪A2)=P(A1)+P(A2) ?
P(A_1 \cup A_2) = P(A_1)+ P(A_2)~?
P(A1∪A2)=kk1+k2=kk1+kk2
P(A_1 \cup A_2) = \frac{k_1 + k_2}{k} = \frac{k_1}{k} + \frac{k_2}{k}
=P(A1)+P(A2)
= P(A_1)+ P(A_2)
A1
A_1
A2
A_2
Probability as relative frequency
Does P() satisfy the axioms?
Ω
\Omega
P(A1∪A2)=P(A1)+P(A2) ?
P(A_1 \cup A_2) = P(A_1)+ P(A_2)~?
∣Ω∣=n⟹2n subsets⟹2n events
|\Omega| = n \implies 2^n~subsets \implies 2^n~events
A1,A2,A3,…,An
A_1, A_2, A_3, \dots, A_n
Suppose are the outcomes
Every event is a union of these outcomes
If the frequencies of are known then the probability of any event can be computed
A1,A2,A3,…,An
A_1, A_2, A_3, \dots, A_n
(axioms are about events)
Examples
A dataset contains images of beaches (60000), mountains (25000) and forests (15000)
What is the probability that a randomly picked image would be of a forest?
Experiment: Select an image
Number of trials: 100000
Frequency of the event "forest": 15000
P(forest)=10000015000=0.15
P(forest) = \frac{15000}{100000} = 0.15
Examples
A country tests 20 million randomly selected people and finds that 1 million are infected
What is the probability that a randomly picked person would be infected?
Experiment: Perform a test
Number of trials: 20 million
Frequency of the event "infected": 1 million
P(infected)=200000001000000=0.05
P(infected) = \frac{1000000}{20000000} = 0.05
Examples
By May-10-2020, India had tested 1673688 samples of which 67176 were found to be positive. Does this mean the probability that a randomly selected person being infected is 0.04
A subtle point: the sample from which the probabilities were estimated should be drawn from the same population on which we are interested in making inferences
No: testing in India was not random but only for people with flu-like symptoms
Flu
Flu
Ω
\Omega
Designing Probability Functions
(the case of equally likely outcomes)
Equally likely outcomes
P(H)=P(T)=k
P(H) = P(T) = k
Ω=H∪T
\Omega = H \cup T
P(Ω)=P(H∪T)
P(\Omega) = P(H \cup T)
=P(H)+P(T)
= P(H) + P(T)
Ω={H,T}
\Omega = \{H, T\}
H
H
T
T
=2k
= 2k
=1
= 1
∴P(H)=P(T)=k=21
\therefore P(H) = P(T) = k = \frac{1}{2}
We can now compute the probability of all 4 subsets of
Ω
\Omega
ϕ,{H},{T},{H,T}
\phi, \{H\}, \{T\}, \{H, T\}
Equally likely outcomes
Ai
A_i
: event that the out come is i
A1,A2,A3,A4,A5,A6
A_1, A_2, A_3, A_4, A_5, A_6
partition
Ω
\Omega
P(A1)=P(A2)=P(A3)=P(A4)=P(A5)=P(A6)=k
P(A_1) = P(A_2) = P(A_3) = P(A_4) = P(A_5) = P(A_6) = k
P(S)=∑i=16P(Ai)=6k=1
P(S) = \sum_{i=1}^6 P(A_i) = 6k = 1
∴P(Ai)=61
\therefore P(A_i) = \frac{1}{6}
We can now compute the probability of all subsets of
Ω
\Omega
Ω={1,2,3,4,5,6}
\Omega = \{1, 2, 3, 4, 5, 6\}
A1
A_1
A2
A_2
A3
A_3
A4
A_4
A5
A_5
A6
A_6
Equally likely outcomes
We can now compute the probability of all subsets of
Ω
\Omega
E
E
: outcome is even
Ω={1,2,3,4,5,6}
\Omega = \{1, 2, 3, 4, 5, 6\}
A1
A_1
A2
A_2
A3
A_3
A4
A_4
A5
A_5
A6
A_6
O
O
: outcome is odd
D
D
: outcome is divisible by 3
Equally likely outcomes
Can we derive a formula for computing the probability of events of an experiment with n equally likely outcomes?
E
E
: any event with k outcomes
(the outcomes are of course disjoint)
P(E)=∑i=1kn1=nk
P(E) = \sum_{i=1}^{k} \frac{1}{n} = \frac{k}{n}
n1
\frac{1}{n}
P(X)=number of outcomes in Ωnumber of outcomes in X
P(X) = \frac{number~of~outcomes~in~X}{number~of~outcomes~in~\Omega}
Equally likely outcomes
Are the axioms of probability satisfied?
n1
\frac{1}{n}
P(X)=number of outcomes in Ωnumber of outcomes in X
P(X) = \frac{number~of~outcomes~in~X}{number~of~outcomes~in~\Omega}
P(Ai)≥0 ?
P(A_i) \geq 0~?
: ratio of two positive numbers
P(Ω)=1 ?
P(\Omega)=1~?
P(A1∪A2)=P(A1)+P(A2) ?
P(A_1 \cup A_2) = P(A_1)+ P(A_2)~?
: contains all outcomes
A1
A_1
A2
A_2
P(A1∪A2)=nk1+k2=nk1+nk2
P(A_1 \cup A_2) = \frac{k_1 + k_2}{n} = \frac{k_1}{n} + \frac{k_2}{n}
=P(A1)+P(A2)
= P(A_1) + P(A_2)
Examples
What is the probability of getting a black card?
n1
\frac{1}{n}
P(B)=5226
P(B) = \frac{26}{52}
P(X)=number of outcomes in Ωnumber of outcomes in X
P(X) = \frac{number~of~outcomes~in~X}{number~of~outcomes~in~\Omega}
What is the probability of getting 3 aces?
n=(352)=22100
n = {52 \choose 3} = 22100
P(A)=221004
P(A) = \frac{4}{22100}
Examples
What is the probability of hitting the red circle at the centre?
P(C)=πR2πr2=(Rr)2
P(C) = \frac{\pi r^2}{\pi R^2} = (\frac{r}{R})^2
R
R
: radius of dartboard
r
r
: radius of red circle
Summary
Learning Objectives
What are sets and some of their properties
What are experiments, sample spaces, outcomes and events?
What are the axioms of probability?
What are some simple ways of defining a probability function?
What are some important theorems: multiplication rule, total probability theorem and Bayes' theorem?
What are independent events?
Set Theory
Finite, Countably infinite, Uncountably infinite
Intersection, Union, Complement
Properties of set operations
Disjoint sets
Axioms of Probability
Designing Probability Functions
Goal: Assign a number to each event such that this number reflects the chance of the experiment resulting in that event
Required: The probability function must satisfy the axioms of probability
Probability as long term relative frequency
Equally likely outcomes
Conditional Probabilities
Change in belief
Before start of play: What is the chance of India winning?
(assume fair playing conditions & equally good teams)
India scores 395 batting first: What is the chance of India winning?
>0.5
> 0.5
0.5
0.5
Change in belief
What exactly happened here?
(assume fair playing conditions & equally good teams)
A: event that India will win
B: India scores 395 runs
P(A) changes once we know that event B has occurred
P(A∣B)
P(A|B)
=P(A)
\neq P(A)
Change in belief
What is the probability that a randomly selected person is healthy (not infected)?
A: event that a person is healthy
B: event that the person has COVID-19 symptoms
10% of the population is infected
P(A)=0.9
P(A) = 0.9
P(A∣B)=P(A)
P(A|B) \neq P(A)
is called the conditional probability of the event A given the event B
P(A∣B)
P(A|B)
The definition of P(B|A)
P(A)=365
P(A) = \frac{5}{36}
| (1 , 1) | (1 , 2) | (1 , 3) | (1 , 4) | (1 , 5) | (1 , 6) |
|---|---|---|---|---|---|
| (2, 1) | (2, 2) | (2, 3) | (2, 4) | (2, 5) | (2, 6) |
| (3, 1) | (3, 2) | (3, 3) | (3, 4) | (3, 5) | (3, 6) |
| (4, 1) | (4, 2) | (4, 3) | (4, 4) | (4, 5) | (4, 6) |
| (5, 1) | (5, 2) | (5, 3) | (5, 4) | (5, 5) | (5, 6) |
| (6, 1) | (6, 2) | (6, 3) | (6, 4) | (6, 5) | (6, 6) |
What is the probability that the sum is 8?
The definition of P(B|A)
P(A∣B)=61
P(A|B) = \frac{1}{6}
| (1 , 1) | (1 , 2) | (1 , 3) | (1 , 4) | (1 , 5) | (1 , 6) |
|---|---|---|---|---|---|
| (2, 1) | (2, 2) | (2, 3) | (2, 4) | (2, 5) | (2, 6) |
| (3, 1) | (3, 2) | (3, 3) | (3, 4) | (3, 5) | (3, 6) |
| (4, 1) | (4, 2) | (4, 3) | (4, 4) | (4, 5) | (4, 6) |
| (5, 1) | (5, 2) | (5, 3) | (5, 4) | (5, 5) | (5, 6) |
| (6, 1) | (6, 2) | (6, 3) | (6, 4) | (6, 5) | (6, 6) |
What is the probability that the sum is 8 given that the first dice shows a 4?
A: sum is 8
B: first dice shows a 4
The definition of P(B|A)
P(B)=366
P(B) = \frac{6}{36}
A: sum is 8
B: first dice shows a 4
Ω
\Omega
(6,2)
(6, 2)
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(1, 1)~(1,2)~(1,3)~(1,4)~(1,5)~(1,6)
(4,4)
(4, 4)
(3,5)
(3, 5)
(5,3)
(5, 3)
(4,1) (4,2) (4,3) (4,5) (4,6)
(4, 1)~~(4,2)~~(4,3)~~(4,5)~~(4,6)
(2,6)
(2, 6)
(2,1) (2,2) (2,3) (2,4) (2,5)
(2, 1)~(2,2)~(2,3)~(2,4)~(2,5)
(3,1) (3,2) (3,3) (3,4) (3,6)
(3, 1)~(3,2)~(3,3)~(3,4)~(3,6)
(5,1) (5,2) (5,4) (5,5) (5,6)
(5, 1)~(5,2)~(5,4)~(5,5)~(5,6)
(6,1) (6,3) (6,4) (6,5) (6,6)
(6, 1)~(6,3)~(6,4)~(6,5)~(6,6)
A
A
B
B
A∩B
A \cap B
P(A∩B)=361
P(A\cap B) = \frac{1}{36}
P(A∣B)=P(B)P(A∩B)=366361=61
P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{\frac{1}{36}}{\frac{6}{36}} = \frac{1}{6}
is called the conditional probability of the event A given the event B
P(A∣B)
P(A|B)
P(A∣B)=P(B)P(A∩B)
P(A|B) = \frac{P(A\cap B)}{P(B)}
conditional probability
regular probabilities
(we already know how to compute these)
Examples
I am thinking of a two digit number. Suppose I tell you that at least one of the two digits is even then what is the probability that both are even?
A: event that both digits are even
B: event that at least one digit is even
P(A)=9020=92
P(A) = \frac{20}{90} = \frac{2}{9}
10, 11, 12, 13, 14, 15, ...., ..., ..., ..., ..., ..., ..., ..., ..., 94, 95, 96, 97, 98, 99
(all equally likely)
but we are interested in P(A|B)
Examples
I am thinking of a two digit number. Suppose I tell you that at least one of the two digits is even then what is the probability that both are even?
A: event that both digits are even
B: event that at least one digit is even
P(A∣B)=P(B)P(A∩B)=90659020=134
P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{20}{90}}{\frac{65}{90}} = \frac{4}{13}
10, 11, 12, 13, 14, 15, ...., ..., ..., ..., ..., ..., ..., ..., ..., 94, 95, 96, 97, 98, 99
(all equally likely)
A
A
B
B
Examples
60% of the students in a class opt for ML. 20% of the students opt for both ML and DL. Given that a student has opted for ML what is the probability that she has also opted for DL?
A: event that student has opted for DL
B: event that the student has opted for ML
P(A∣B)=P(B)P(A∩B)=0.60.2=31
P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.6} = \frac{1}{3}
A
A
B
B
Ω
\Omega
Do conditional probabilities satisfy the axioms of probability?
Axioms of Probability
P(A∣B)=P(B)P(A∩B)≥0
P(A|B) = \frac{P(A\cap B)}{P(B)} \geq 0
: ratio of two probabilities
P(Ω∣A)=P(B)P(Ω∩B)=P(B)P(B)=1
P(\Omega|A)= \frac{P(\Omega \cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1
P(A1∪A2∣B)=P(B)P((A1∪A2)∩B)
P(A_1 \cup A_2|B) = \frac{P((A_1 \cup A_2) \cap B)}{P(B)}
=P(B)P(A1∩B)+P(B)P(A2∩B)
= \frac{P(A_1 \cap B)}{P(B)} + \frac{P(A_2 \cap B)}{P(B)}
=P(A1∣B)+P(A2∣B)
= P(A_1|B) + P(A_2|B)
B
B
A1
A_1
Ω
\Omega
A2
A_2
=P(B)P((A1∩B)∪(A2∩B)
= \frac{P((A_1 \cap B) \cup (A_2 \cap B)}{P(B)}
The multiplication principle
The chain rule of probability
P(A∣B)=P(B)P(A∩B)
P(A|B) = \frac{P(A\cap B)}{P(B)}
P(B∣A)=P(A)P(B∩A)
P(B|A) = \frac{P(B\cap A)}{P(A)}
∴P(A∩B)=P(A∣B)⋅P(B)
\therefore P(A\cap B) = P(A|B)\cdot P(B)
∴P(B∩A)=P(B∣A)⋅P(A)
\therefore P(B\cap A) = P(B|A)\cdot P(A)
∴P(A∩B)=P(A∣B)⋅P(B)=P(B∣A)⋅P(A)
\therefore P(A\cap B) = P(A|B)\cdot P(B) = P(B|A)\cdot P(A)
The chain rule of probability
+
+
B
B
Ω
\Omega
A
A
A: event that a person is infected
B: event that the test result is positive
A∩B
A\cap B
A∩Bc
A\cap B^\mathsf{c}
Ac∩B
A^\mathsf{c}\cap B
Ac∩Bc
A^\mathsf{c}\cap B^\mathsf{c}
The chain rule of probability
+
+
B
B
Ω
\Omega
A
A
roughly 10% of the population is infected
for an infected person the test shows a negative result in 1% of the cases
Facts:
P(A)=0.1
P(A) = 0.1
P(Bc∣A)=0.01
P(B^\mathsf{c}|A) = 0.01
⟹P(B∣A)=0.99
\implies P(B|A) = 0.99
⟹P(Bc∣Ac)=0.95
\implies P(B^\mathsf{c}|A^\mathsf{c}) = 0.95
P(B∣Ac)=0.05
P(B|A^\mathsf{c}) = 0.05
for an healthy person the test shows a positive result in 5% of the cases
The chain rule of probability
+
+
B
B
Ω
\Omega
A
A
Facts:
P(A)=0.1
P(A) = 0.1
P(Bc∣A)=0.01
P(B^\mathsf{c}|A) = 0.01
⟹P(B∣A)=0.99
\implies P(B|A) = 0.99
⟹P(Bc∣Ac)=0.95
\implies P(B^\mathsf{c}|A^\mathsf{c}) = 0.95
P(B∣Ac)=0.05
P(B|A^\mathsf{c}) = 0.05
A
A
A∩B
A \cap B
A∩Bc
A \cap B^\mathsf{c}
Ac∩B
A^\mathsf{c} \cap B
Ac∩Bc
A^\mathsf{c} \cap B^\mathsf{c}
Ac
A^\mathsf{c}
B∣A
B|A
Bc∣A
B^\mathsf{c}|A
B∣Ac
B|A^\mathsf{c}
Bc∣Ac
B^\mathsf{c}|A^\mathsf{c}
A: infected
B: test positive
The chain rule of probability
for n events
P(A∩B∩C)=P((A∩B)∩C)
P(A \cap B \cap C) = P((A \cap B) \cap C)
Let (A∩B)=X
Let~(A \cap B)=X
∴P(A∩B∩C)=P(X∩C)
\therefore P(A \cap B \cap C) = P(X \cap C)
∴P(A∩B∩C)=P(X)⋅P(C∣X)
\therefore P(A \cap B \cap C) = P(X)\cdot P(C|X)
∴P(A∩B∩C)=P(A∩B)⋅P(C∣A∩B)
\therefore P(A \cap B \cap C) = P(A\cap B)\cdot P(C|A \cap B)
∴P(A∩B∩C)=P(A)⋅P(B∣A)⋅P(C∣A∩B)
\therefore P(A \cap B \cap C) = P(A)\cdot P(B|A)\cdot P(C|A \cap B)
The chain rule of probability
for n events
P(A∩B∩C∩D)
P(A \cap B \cap C \cap D)
=P(A)⋅P(B∣A)⋅P(C∣A∩B)⋅P(D∣A∩B∩C)
= P(A)\cdot P(B|A)\cdot P(C|A \cap B)\cdot P(D|A \cap B \cap C)
P(A1∩A2∩⋯∩An)
P(A_1 \cap A_2 \cap \dots \cap A_n)
=P(A1)∏i=2nP(Ai∣A1…Ai−1)
= P(A_1)\prod_{i=2}^{n} P(A_i|A_1\dots A_{i-1})
Examples
Suppose you draw 3 cards one by one without replacement. What is the probability that all the 3 cards are aces?
Using counting principles:
p=(352)(34)=49! 3!52!1! 3!4!=52∗51∗504∗3∗2
p = \frac{4 \choose 3}{52 \choose 3} = \frac{\frac{4!}{1!~3!}}{\frac{52!}{49!~3!}} = \frac{4*3*2}{52*51*50}
Examples
Suppose you draw 3 cards one by one without replacement. What is the probability that all the 3 cards are aces?
Using chain rule:
Ai
A_i
:the event that the i-th card is an ace
P(A1∩A2∩A3)
P(A_1\cap A_2\cap A_3)
=P(A1)⋅P(A2∣A1)⋅P(A3∣A2∩A1)
= P(A_1)\cdot P(A_2|A_1)\cdot P(A_3|A_2 \cap A_1)
P(A1)
P(A_1)
=524
= \frac{4}{52}
P(A2∣A1)
P(A_2|A_1)
=513
= \frac{3}{51}
P(A3∣A1∩A2)
P(A_3|A_1\cap A_2)
=502
= \frac{2}{50}
=52∗51∗504∗3∗2
= \frac{4*3*2}{52*51*50}
Total Probability Theorem
Total Probability Theorem
A1
A_1
A5
A_5
A4
A_4
A3
A_3
A2
A_2
A6
A_6
A7
A_7
B
B
A1,A2,⋯An
A_1, A_2, \cdots A_n
Ω
\Omega
partition
Ω
\Omega
A1∪A2,∪⋯∪An=Ω
A_1 \cup A_2, \cup \dots \cup A_n = \Omega
Ai∩Aj=ϕ ∀i=j
A_i \cap A_j = \phi~\forall i \neq j
B=(B∩A1)∪(B∩A2)∪⋯∪(B∩An)
B = (B \cap A_1) \cup (B \cap A_2) \cup \dots \cup (B \cap A_n)
P(B)=P(B∩A1)+P(B∩A2)+⋯+P(B∩An)
P(B) = P(B \cap A_1) + P(B \cap A_2) + \dots + P(B \cap A_n)
P(B)=∑i=1nP(Ai)⋅P(B∣Ai)
P(B) = \sum_{i=1}^{n} P (A_i) \cdot P(B | A_i)
P(B)
P(B)
=P(A1)⋅P(B∣A1)+P(A2)⋅P(B∣A2)+⋯+P(An)⋅P(B∣An)
= P(A_1)\cdot P(B|A_1) + P(A_2)\cdot P(B|A_2) + \dots + P(A_n)\cdot P(B|A_n)
Examples
+
+
B
B
Ω
\Omega
A
A
Facts:
P(A)=0.1
P(A) = 0.1
P(Bc∣A)=0.01
P(B^\mathsf{c}|A) = 0.01
⟹P(B∣A)=0.99
\implies P(B|A) = 0.99
⟹P(Bc∣Ac)=0.95
\implies P(B^\mathsf{c}|A^\mathsf{c}) = 0.95
P(B∣Ac)=0.05
P(B|A^\mathsf{c}) = 0.05
A: infected
B: test positive
P(B)=?
P(B) = ?
P(B)=P(A)P(B∣A)+P(Ac)(B(∣Ac)
P(B) = P(A)P(B|A) + P(A^\mathsf{c})(B(|A^\mathsf{c})
∴P(B)=0.1∗0.99+0.9∗0.05=0.144
\therefore P(B) = 0.1*0.99 + 0.9*0.05 =0.144
Examples
What is the probability that he will come out alive?
P(A1)=P(A2)=P(A3)=31
P(A_1)=P(A_2)=P(A_3) = \frac{1}{3}
P(B∣A1)=0.3
P(B|A_1) = 0.3
P(B∣A2)=0.6
P(B|A_2) = 0.6
P(B∣A3)=0.75
P(B|A_3) = 0.75
P(Bc)=?
P(B^\mathsf{c}) = ?
B : monster encountered
i-th path taken
Ai:
A_i:
=P(A1)P(Bc∣A1)+P(A2)P(Bc∣A2)+P(A3)P(Bc∣A3)
= P(A_1)P(B^\mathsf{c}|A_1) + P(A_2)P(B^\mathsf{c}|A_2) + P(A_3)P(B^\mathsf{c}|A_3)
=31⋅0.7+31⋅0.4+31⋅0.25=0.45
= \frac{1}{3}\cdot0.7 + \frac{1}{3}\cdot0.4 + \frac{1}{3}\cdot0.25 = 0.45
Bayes' Theorem
Examples
If he does not come out alive what is the probability that he took path A1?
P(A1∣B)=?
P(A_1|B) = ?
B : monster encountered
i-th path taken
Ai:
A_i:
P(A1∣B)=P(B)P(A1∩B)
P(A_1|B) = \frac{P(A_1 \cap B)}{P(B)}
P(A1∣B)=P(A1)⋅P(B∣A1)+P(A2)⋅P(B∣A2)+P(A3)⋅P(B∣A3)P(A1∩B)
P(A_1|B) = \frac{P(A_1 \cap B)}{P(A_1)\cdot P(B|A_1) + P(A_2)\cdot P(B|A_2) + P(A_3)\cdot P(B|A_3)}
P(A1∩B)=P(A1)P(B∣A1)P(A1∩B)=P(B)P(A1∣B)
P(A_1 \cap B) = P(A_1)P(B|A_1) \\
P(A_1 \cap B) = P(B)P(A_1|B)
P(A1)=P(A2)=P(A3)=31
P(A_1)=P(A_2)=P(A_3) = \frac{1}{3}
P(B∣A1)=0.3
P(B|A_1) = 0.3
P(B∣A2)=0.6
P(B|A_2) = 0.6
P(B∣A3)=0.75
P(B|A_3) = 0.75
Examples
If he does not come out alive what is the probability that he took path A1?
P(A1)=P(A2)=P(A3)=31
P(A_1)=P(A_2)=P(A_3) = \frac{1}{3}
P(B∣A1)=0.3
P(B|A_1) = 0.3
P(B∣A2)=0.6
P(B|A_2) = 0.6
P(B∣A3)=0.75
P(B|A_3) = 0.75
P(A1∣B)=?
P(A_1|B) = ?
B : monster encountered
i-th path taken
Ai:
A_i:
P(A1∣B)=P(B)P(A1∩B)
P(A_1|B) = \frac{P(A_1 \cap B)}{P(B)}
P(A1∣B)=P(A1)⋅P(B∣A1)+P(A2)⋅P(B∣A2)+P(A3)⋅P(B∣A3)P(A1)P(B∣A1)
P(A_1|B) = \frac{P(A_1)P(B|A_1)}{P(A_1)\cdot P(B|A_1) + P(A_2)\cdot P(B|A_2) + P(A_3)\cdot P(B|A_3)}
P(A1∩B)=P(A1)P(B∣A1)P(A1∩B)=P(B)P(A1∣B)
P(A_1 \cap B) = P(A_1)P(B|A_1) \\
P(A_1 \cap B) = P(B)P(A_1|B)
=0.182
= 0.182
Examples
If he does not come out alive what is the probability that he took path A3?
P(A1)=P(A2)=P(A3)=31
P(A_1)=P(A_2)=P(A_3) = \frac{1}{3}
P(B∣A1)=0.3
P(B|A_1) = 0.3
P(B∣A2)=0.6
P(B|A_2) = 0.6
P(B∣A3)=0.75
P(B|A_3) = 0.75
P(A3∣B)=?
P(A_3|B) = ?
B : monster encountered
i-th path taken
Ai:
A_i:
P(A3∣B)=P(B)P(A3∩B)
P(A_3|B) = \frac{P(A_3 \cap B)}{P(B)}
P(A3∣B)=P(A1)⋅P(B∣A1)+P(A2)⋅P(B∣A2)+P(A3)⋅P(B∣A3)P(A3)P(B∣A3)
P(A_3|B) = \frac{P(A_3)P(B|A_3)}{P(A_1)\cdot P(B|A_1) + P(A_2)\cdot P(B|A_2) + P(A_3)\cdot P(B|A_3)}
P(A3∩B)=P(A3)P(B∣A3)P(A3∩B)=P(B)P(A3∣B)
P(A_3 \cap B) = P(A_3)P(B|A_3) \\
P(A_3 \cap B) = P(B)P(A_3|B)
=0.45
= 0.45
Breaking it down
Exploit Multiplication Rule
P(A1)P(B∣A1)=P(B)P(A1∣B)
P(A_1)P(B|A_1) = P(B)P(A_1|B)
Exploit Total Probability Theorem
P(B)=P(A1)⋅P(B∣A1)+P(A2)⋅P(B∣A2)+P(A3)⋅P(B∣A3)
P(B) = P(A_1)\cdot P(B|A_1) + P(A_2)\cdot P(B|A_2) + P(A_3)\cdot P(B|A_3)
Exploit the known probabilities
P(A1∣B)=∑i=1nP(Ai)P(B∣Ai)P(A1)⋅P(B∣A1)
P(A_1|B) = \frac{P(A_1)\cdot P(B|A_1) }{\sum_{i=1}^nP(A_i)P(B|A_i)}
Bayes' Theorem
Breaking it down: Example 1
A : Ship 1 sends a signal 1
B : Ship 2 receives a signal 1
P(A)=0.01
P(A) = 0.01
P(B∣A)=0.95
P(B|A) = 0.95
P(B∣Ac)=0.05
P(B|A^\mathsf{c}) = 0.05
P(A∣B)=?
P(A|B) = ?
P(A∣B)=P(A)P(B∣A)+P(Ac)P(B∣Ac)P(A)P(B∣A)=0.18
P(A|B) = \frac{P(A)P(B|A)}{P(A)P(B|A) + P(A^\mathsf{c})P(B|A^\mathsf{c})} = 0.18
A
A
Ac
A^\mathsf{c}
10000
10000
9900
9900
100
100
495
495
95
95
Breaking it down: Example 2
A
A
Ac
A^\mathsf{c}
10000
10000
9000
9000
1000
1000
450
450
990
990
+
+
B
B
Ω
\Omega
A
A
Facts:
P(A)=0.1
P(A) = 0.1
P(Bc∣A)=0.01
P(B^\mathsf{c}|A) = 0.01
⟹P(B∣A)=0.99
\implies P(B|A) = 0.99
⟹P(Bc∣Ac)=0.95
\implies P(B^\mathsf{c}|A^\mathsf{c}) = 0.95
P(B∣Ac)=0.05
P(B|A^\mathsf{c}) = 0.05
A: infected
B: test positive
P(A∣B)=?
P(A|B) = ?
P(A∣B)=P(A)P(B∣A)+P(Ac)P(B∣Ac)P(A)P(B∣A)=0.6875
P(A|B) = \frac{P(A)P(B|A)}{P(A)P(B|A) + P(A^\mathsf{c})P(B|A^\mathsf{c})} = 0.6875
Bayes' Theorem: 3 forms
P(Ai∣B)=∑j=1nP(Aj)P(B∣Aj)P(Ai)⋅P(B∣Ai)
P(A_i|B) = \frac{P(A_i)\cdot P(B|A_i) }{\sum_{j=1}^{n}P(A_j)P(B|A_j)}
P(A∣B)=P(B)P(A∩B)
P(A|B) = \frac{P(A \cap B)}{P(B)}
P(A∣B)=P(B)P(A)⋅P(B∣A)
P(A|B) = \frac{P(A)\cdot P(B|A) }{P(B)}
Independent Events
Do we always update our beliefs
A: I had a sandwich for breakfast
If A occurs will you update your belief about B ?
What do you call such events?
B: It will rain today
No
Independent events
Example
50 girls and 70 boys in a class. Of these, 35 girls and 49 boys are good at Maths. If I tell you that a student is very good at Maths what is the probability that she is a girl?
Facts:
P(A)=50+7050=125
P(A) = \frac{50}{50+70} = \frac{5}{12}
P(B∣A)=5035=107
P(B|A) = \frac{35}{50} = \frac{7}{10}
A: student is girl
B: student is good at Maths
P(Ac)=127
P(A^\mathsf{c}) = \frac{7}{12}
P(B∣Ac)=7049=107
P(B|A^\mathsf{c}) = \frac{49}{70} = \frac{7}{10}
P(A∣B)=?
P(A|B) = ?
P(A∣B)=P(B)P(B∣A)P(A)
P(A|B) = \frac{P(B|A)P(A)}{P(B)}
P(B)=P(B∣A)P(A)+P(B∣A′)P(A′)=107
P(B) = P(B|A)P(A) + P(B|A')P(A') = \frac{7}{10}
=107107125=125
= \frac{\frac{7}{10}\frac{5}{12}}{\frac{7}{10}} = \frac{5}{12}
Tow events A and B are independent if
P(A∣B)=P(B)
P(A|B) = P(B)
P(A∩B)=P(B)⋅P(A∣B)
P(A \cap B) = P(B)\cdot P(A|B)
Tow events A and B are independent if
P(A∩B)=P(A)⋅P(B)
P(A\cap B) = P(A)\cdot P(B)
Example
A: first toss results in a head
B: exactly 2 tosses result in heads
P(A)=84
P(A) = \frac{4}{8}
Are A and B independent ?
| H | H | H | * | ||
| H | H | T | * | * | * |
| H | T | H | * | * | * |
| H | T | T | * | ||
| T | H | H | * | ||
| T | H | T | |||
| T | T | H | |||
| T | T | T |
A
A
B
B
A∩B
A \cap B
P(B)=83
P(B) = \frac{3}{8}
P(A∩B)=82
P(A \cap B) = \frac{2}{8}
P(A∩B)=P(A)P(B)
P(A \cap B) \neq P(A)P(B)
Example
Are A and B independent ?
A: sum is 7
B: second dice shows an even number
A={(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}
A = \{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) \}
B={(1,2),(1,4),(1,6),(2,2),(2,4),(2,6),(3,2),(3,4),(3,6),(4,2),(4,4),(4,6),(5,2),(5,4),(5,6),(6,2),(6,4),(6,6)}
B = \{(1,2), (1,4), (1,6), (2,2), (2,4), (2,6),(3,2), (3,4), (3,6), \newline(4,2), (4,4), (4,6), (5,2), (5,4), (5,6), (6,2), (6,4), (6,6)\}
A∩B={(1,6),(3,4),(5,2)}
A \cap B = \{(1,6), (3,4), (5,2) \}
P(A)=366=61
P(A) = \frac{6}{36} = \frac{1}{6}
P(A∩B)=363=121
P(A \cap B) = \frac{3}{36}= \frac{1}{12}
P(B)=3618=21
P(B) = \frac{18}{36} = \frac{1}{2}
∴P(A∩B)=P(A)⋅P(B)
\therefore P(A \cap B) = P(A)\cdot P(B)
Example
A: first answer is correct
B: second answer is correct
A quiz has two multiple choice Qs. The first Q has 4 choices of which 1 is correct and the second Q has 3 choices of which 1 is correct. If a student randomly guesses the answers what is the probability that he will answer both Qs correctly?
P(A)=41
P(A) = \frac{1}{4}
P(B)=31
P(B) = \frac{1}{3}
P(A∩B)=P(A)⋅P(B)=121
P(A \cap B) = P(A)\cdot P(B) = \frac{1}{12}
Independence: n events
We say that events are pairwise independent if
A1,A2,A3,…,An
A_1, A_2, A_3, \dots, A_n
P(Ai∩Aj)=P(Ai)⋅P(Aj) ∀ i=j
P(A_i \cap A_j) = P(A_i)\cdot P(A_j)~\forall~i\neq j
We say that events are mutually independent or independent if for all subsets
P(∩i∈IAi)=∏i=1nP(Ai)
P(\cap_{i \in I} A_i) = \prod_{i=1}^{n}P(A_i)
A1,A2,A3,…,An
A_1, A_2, A_3, \dots, A_n
I⊂{1,2,3,…,n}
I \subset \{1,2,3,\dots,n\}
{1,2,3}
\{1,2,3\}
n=3
n = 3
{1,2},{1,3},{2,3},{1,2,3}
\{1,2\}, \{1, 3\}, \{2,3\}, \{1, 2, 3\}
P(A1∩A2)=P(A1)⋅P(A2)
P(A_1 \cap A_2 ) = P(A_1)\cdot P(A_2)
P(A1∩A3)=P(A1)⋅P(A3)
P(A_1 \cap A_3 ) = P(A_1)\cdot P(A_3)
P(A2∩A3)=P(A2)⋅P(A3)
P(A_2 \cap A_3 ) = P(A_2)\cdot P(A_3)
P(A1∩A2∩A3)
P(A_1 \cap A_2 \cap A_3 )
=P(A1)⋅P(A2)⋅P(A3)
= P(A_1)\cdot P(A_2)\cdot P(A_3)
Summary
Putting it all together
Conditional Probability
Compute P(B∣A) using P(A∩B) and P(A)
Compute~P(B|A)~using~P(A\cap B)~and~P(A)
Multiplication Rule
Compute P(A∩B) using P(A) and P(B∣A)
Compute~P(A\cap B)~using~P(A)~and~P(B|A)
Total Probability Theorem
Compute P(B) using P(A1),P(A2),…,P(An)
Compute~P(B)~using~P(A_1), P(A_2), \dots, P(A_n)
and P(B∣A1),P(B∣A2),…,P(B∣An)
and~P(B|A_1), P(B|A_2), \dots, P(B|A_n)
Bayes' Theorem
Compute P(A∣B) using
Compute~P(A|B)~using
P(∩i∈IAi)=∏i=1nP(Ai)
P(\cap_{i \in I} A_i) = \prod_{i=1}^{n}P(A_i)
Independent events
Recap: List of Topics Descriptive Statistics Probability Theory Inferential Statistics Different types of data Different types of plots Measures of centrality and spread Counting, Sample spaces, events Discrete and continuous RVs Bernoulli, Uniform, Normal dist. Sampling strategies Interval Estimators Hypothesis testing (z-test, t-test) ANOVA, Chi-square test Linear Regression
FDS_Intro_To_Probability
By One Fourth Labs
FDS_Intro_To_Probability
PadhAI One: FDS Week 3 (MK)
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