GEM2900 Cheat Sheet 1

Chapter 1: Interpretations of Probability Chance experiment: a process with uncertainty about resulted outcomes, e.g. drawing a card from a wellshuffled deck Event: collection of possible outcomes from a chance experiment, e.g. event where card drawn is an Ace

Long-run relative frequency For an event to have P=1/4, we expect it to arise one-quarter of the time. It doesn’t necessarily occur once in every block of 4 repetitions. Long-run frequency interpretations is limited to repeated conditions HHHHHHHT & HTTHHTTHT just as likely to occur

Degrees of Belief

p 1−p ∧odds against = 1−p p

E.g. given event A and

P( B) P ( A ∩ B ) =P ( A )¿

Dutch Book – A set of odds and bets which guarantees a profit (or loss) regardless of the outcomes of the gamble. Associated with incoherent probabilities and appropriate arrangement of bets

Prosecutor’s fallacy

P ( A|B ) ≠ P(B∨ A)

Multiplication Rule,

P ( A ∩ B ) =P ( B| A ) P ( A )

Event A and A sample space

covers entire

Whenever AB occurs, A (or B) must occur The composition of two events is always less probable than each individual event Events A and B are mutually exclusive if they don’t intercept Mutually ExclusiveIndependent unless probability of one is 0

c

P ( A )+ P ( A )=1 P ( AB ) ≤ P ( A ) P ( AB ) ≤ P ( B ) P ( A ∩ B ) =0 If A ∩B=∅→ P ( A ∪ B )=P ( A ) + P ( B )

Positively associated

P ( A|B )=

P ( B| A ) P ( A ) P ( B| A ) P ( A )+ P ( B| A C ) P( A C )

Has disease Sensitivity

No disease Type I error (false +ve) Specificity

Chapter 4: Computing Probabilities

If first three equations are held, A, B, and, C are pairwise

P ( A ) ≠ 0, P ( B ) ≠ 0

Bayes’ Theorem

Negativ Type II error (false – e ve) Better to have high sensitivity than high specificity, safe than sorry

is 2 to 1

Chapter 2 + 3: The Rule Book + Conditional Probability

i=1

Positive

independent Probability of A given B

C

n

P ( B ) =∑ P ( B| A i ) P( A i)

Specificity VS Sensitivity

P(C) P ( A ∩C )=P( A) ¿

P (B) P(C) P ( A ∩ B ∩C )=P( A) ¿

, odds in favour of A

P ( B ) =P ( B ∩ A ) + P ( B ∩ A C ) ¿ P ( B| A ) P ( A ) + P ( B| A C ) P ( A C )

P ( A|B )=P ( A )

P (C) P ( B ∩C )=P( B)¿

Subjective probability OR judgmental probability

2 P ( A )= 3

P ( AB c ) =P ( A ) P( Bc )

Events A, B, and, C are mutually independent IF

Obtaining 6 when rolling a die Obtaining head when tossing a coin

odds∈favour=

When events A and B are independent, so are their complements

Law of Total Probability

P ( B| A )=P( B)

Interpretations of Probabilities: Propensities / Classical e.g .

P ( AB )=P ( A ) P ( B )

P ( A|B )=

P ( A ∩ B) P(B)

¿ P ( A|B ) P(B)

Laplace experiments: outcome equally likely

experiments

each

Factorial Ways of arrangement Permutati Order MATTERS on Combinati Order doesn’t matter on Set Identities Associative ( A ∪B ) ∪ C= A ∪ ( B ∪C ) Laws

( A ∩B ) ∩C= A ∩(B ∩C )

Negatively associated

Distributive P ( A|B )> P ( A ) → P ( B| A ) > P(B) P ( A|B )< P ( A ) → P ( B| A ) < P(B) Laws

Inclusion-Exclusion Formula

P ( A ∪ B )=P ( A ) + P ( B )−P ( A ∩ B )

If events mutually exclusive If events are independent

P ( A ∪ B )=P ( A ) + P ( B )

Absorption Laws

P ( A ∪ B )=P ( A ) + P ( B )−P ( A ) P (B)

( A ∪ B ) ∩C=( A ∩C ) ∪ ( B ∩C ) ( A ∩B)∪ C=( A ∪C)∩( B ∪ C) A ∪( A ∩ B)= A A ∩( A ∪ B)= A

DeMorgan’s Laws

( A ∪ B)c = A c ∩ Bc

¿ P ( A )+ P ( B ) + P ( C )−P ( A ∩ B )−P ( A ∩C )−P ( B ∩C )+ P ( A ∩B ∩C)

( A ∩B)c = A c ∪ Bc

P ( A ∪ B ∪C )

where

Set Difference Laws Game of Craps

i

2

3

4

1 1 1 P (≥ 1 match )= 4 ∙ − 4 ∙ +…+ 4 ∙ 1 4 2 4 ×3 4 4 × 3× 2× 1

A−B=A ∩B c 5

6

7

8

() ()

9

1 0

1 1

1 2

¿ 1−

Let

()

1 1 1 + − 2! 3 ! 4 !

n

the probability that all six sides show up?

6 !∙ 7 2 P ( all sides show up )= 7 6

()

Birthday Problem be the no. of pairs, n

P (≥ 1 match )=∑ (−1 )

n−1

1

∙

1 n!

P ( distinct birthdays∈n )= ¿

Matching Problem 4 vials of blood & 4 name samples, find P of at least one right match

Chance Roll a die 6 times, what is the probability that all six sides show up?

P ( all sides show up )=

6! 66

Roll a die 7 times, what is

365 P n 365n

Inverse Birthday Problem

n people different 364 n P ( ¿ you ) = 365

( )

365 ×364 × … × ( 365−n+1 ) 36 5 n