STAT 302 Intro to Probabilitybouchard/courses/stat302-sp2017-18//files/... · Prop. 3 De Morgan’s...

STAT 302 Intro to Probability

Instructor: Alexandre Bouchard

www.stat.ubc.ca/~bouchard/courses/stat302-sp2017-18/

http://www.stat.ubc.ca/~bouchard/courses/stat302-sp2017-18/

Plan for today:

• Axioms of probability.

• Reliability, continued.

• Probability tree diagram.

• More examples.

Logistics• Let me know if you cannot access website

• Last reminders:

• Office hours: Contact tab.

• Clickers (see links in Syllabus tab)

• Piazza under Contact tab.

• Readings: posted in Schedule tab

• Slides under Files tab

• Additional practice problems: Syllabus tab

http://www.stat.ubc.ca/~bouchard/courses/stat302-fa2015-16/


Logistics

• I will start using clickers today,

• but I am not taking grade into account (yet)

• We cannot have a larger room unfortunately, but keep monitoring the enrolment!



Disclaimer

• Workload and difficulty increases in the second half of semester (continuous probability)

• Make sure you have the time to stay on top of the material

• Good things to review from pre-requisite courses:

• set theory notation

• bivariate integration

Review• Outcome: a scenario, s =

• Sample space S: all the possible scenarios

• Event: a set of outcomes, e.g. E = {s ∈ S : s is red}

• Probability: in the discrete, equally weighted case, P(E) = |E| / |S|

• Properties:

• P(E ⋃ F) = P(E) + P(F) when E & F are disjoint (non-overlapping)

• P(Ec) = 1 - P(E)

Going beyond the equally weighted case

Discrete, equally weighted

P(E) = |E| / |S|

a) 0 ≤ P(E) ≤ 1b) P(S) = 1c) P(E ⋃ F) = P(E) + P(F) if E and F are disjoint

Definition (1):

Properties (2):

Not equally weighted (or not discrete)

Example: soon - redundancy problem

(ex. 11)

General rules of probability (Kolmogorov’s insight)

a) 0 ≤ P(E) ≤ 1b) P(S) = 1c) P(E ⋃ F ⋃ ...) = P(E) + P(F) + ... if E, F, ... are all disjoint

Assume:These are called the axioms of probability

***

Def. 3

Redundancy is not always bad

• Redundant systems:

• create one or more duplicates of an important part of a machine

• as long as one of the copies works, the machine works as a whole

• the machine only fails when both copies break

• Many examples in biology (kidneys), engineering

Redundant server power

supply

Example of a typical reliability problem

• Known:

• Power supply #1 works 60% of the time at delivery


• You also observed that both power supplies work at delivery 40% of the time

• Question: what is the probability that both power supply are broken at delivery?

Ex. 11

Strategy

1.Define a probability space S

2.Define the known information as events, ex.: W1 = power supplies #1 works

3.Define the goal in terms of the probability of an event

4.Use properties 1 and 2 to find that probability (see next 2 slides as well to make your life easier)

Prop. 3 De Morgan’s law: Distributing complements

DeMorgan’s Laws

We have

(ni=1Ei )c =

n

i=1Ec

i

(ni=1Ei )c =

n

i=1Ec

i

Proof: Suppose x is an outcome of (ni=1Ei )

c , then x is not in

(ni=1Ei ). Thus it is not in any of the event Ei , i = 1, ..., n. Hence it

is in Ecifor all i = 1, ..., n. Hence this proves (n

i=1Ei )c

n

i=1Ec

i.

Suppose x is an outcome of ni=1E

c

i, hence it is in Ec

ifor all

i = 1, ..., n. Hence it is in none of the event Ei , i = 1, ..., n. Thus it is(n

i=1Ei )c and we have proven that n

i=1Ec

i (n

i=1Ei )c . The result

(ni=1Ei )

c = ni=1E

c

ifollows.

To prove (ni=1Ei )

c = ni=1E

c

i, we remark that

(ni=1Ec

i )c = ni=1 (E

c

i )c

previous result applied to events {E ci }

= ni=1Ei as (Ec

i )c = Ei .

Thus by taking the complement on both sides, we obtain the result.

AD () Jan. 2010 7 / 8

union complement

intersectioncomplement

P(E ⋃ F) = P(E) + P(F) if E and F are disjoint, i.e. E ⋂ F = ∅

From earlier:

What if: we want the prob. of non-disjoint unions?

= + -

P(E ⋃ F) P(E) P(F) P(E ⋂ F)

Prop. 4 Inclusion-exclusion: Going between unions and intersections

Ex. 11

• Known:



• You also observed that both power supplies work at delivery 40% of the time

• Question: what is the probability that both power supply are broken at delivery?

De Morgan’s law:Distributing complements

DeMorgan’s Laws

We have

(ni=1Ei )c =

n

i=1Ec

i

(ni=1Ei )c =

n

i=1Ec

i

Proof: Suppose x is an outcome of (ni=1Ei )

c , then x is not in

(ni=1Ei ). Thus it is not in any of the event Ei , i = 1, ..., n. Hence it

is in Ecifor all i = 1, ..., n. Hence this proves (n

i=1Ei )c

n

i=1Ec

i.

Suppose x is an outcome of ni=1E

c

i, hence it is in Ec

ifor all

i = 1, ..., n. Hence it is in none of the event Ei , i = 1, ..., n. Thus it is(n

i=1Ei )c and we have proven that n

i=1Ec

i (n

i=1Ei )c . The result

(ni=1Ei )

c = ni=1E

c

ifollows.

To prove (ni=1Ei )

c = ni=1E

c

i, we remark that

(ni=1Ec

i )c = ni=1 (E

c

i )c

previous result applied to events {E ci }

= ni=1Ei as (Ec

i )c = Ei .

Thus by taking the complement on both sides, we obtain the result.

AD () Jan. 2010 7 / 8

union complement

intersectioncomplement

P(E ⋃ F) = P(E) + P(F) if E and F are disjoint, i.e. E ⋂ F = ∅

From earlier:

What if: we want the prob. of non-disjoint unions?

= + -

P(E ⋃ F) P(E) P(F) P(E ⋂ F)

Prop. 4 Inclusion-exclusion:Going between unions and intersections

A. 8%B. 10%C. 12%D. 14%

Partitions and

probability tree diagrams

Def. 4

Partition of an eventF

E1

E2E3

E4

1. The union of the Ei ’s is equal to F: ⋃i Ei = F

2. The Ei ’s are disjoint: if i ≠ j, then Ei ⋂ Ej = ∅

The events Ei form a partition of the event F if:

Note: as consequence of 1, 2 and the axioms of probability, P(F) = P(E1) + P(E2) + P(E3) + P(E4)

Why is this useful?

{

W1

W1c

{s11}

{s10}

{s01}

{s00}

Def. 5

Edges (arrows) coming out of an event are

partitions of that event

Probability tree diagram

Leaves contain a single outcome

Nodes are events,

containing all the other

ones below

Examples of solving discrete probability problems with tree

diagrams

Ex. 12 Probability that n coins are all tails

Recall:

Probability when outcomes are equally

likely:

# of outcomes of interest# of outcomes

P(E) = |E| / |S|

Here: |E| = ?|S| = ?

Ex. 12 Probability that n coins are all tails

Recall:


likely:


P(E) = |E| / |S|

Here: |E| = 1|S| = ?

Ex. 12 Probability that n coins are all tails: finding |S|

*,*,*

H,*,*

T,*,*

H,H,*

H,T,*

T,H,*

T,T,*

H,H,H

H,H,T

H,T,H

H,T,TT,H,HT,H,T

T,T,HT,T,T

depth = n = 3

|S| = 2|E3| = 2x22

= 2depth

|E3| = 2|E2| = 2x2

|E2| = 2|E1| = 2

|E1| = 1

Probability of winning the lottery

Ex. 13

• You pick your number when you buy a ticket

• The lottery company draws at random from an urn containing n numbered balls {1, 2, ..., n} [example: n=5]

• k times [example: k=3]

• without replacement (each number is either picked 0 or 1 time, not more)

Probability of winning the lottery (without replacement)

Ex. 13

• You win if the numbers you picked match those from the draw.

• See example on the right, do you win in this case?

Your picks: 12, 31, 10, 23, 8, 45

Draw result:

a) if order matters, NOb) if order does not matter, YES

Note: most lottery use (b), but let’s do (a) first---it is simpler

Probability of winning the lottery (order matters, without replacement)

Ex. 13a

Recall:


likely:


P(E) = |E| / |S|

Here: |E| = 1|S| = ?

Find |S|:Ex. 13a

A) 243B) 125C) 60D) 15

Find |S|:Ex. 13a

k = number of draws = 3n = number of balls in urn = 5

* * *

2 * *

3 * *

4 * *

5 * *

1 * *

1 2 *

1 3 *

1 4 *

1 5 *

2 1 *

2 3 *

2 4 *

2 5 *

3 1 *

2 1 32 1 4

2 1 5

......

Hint: the probability tree looks like:

A) 243B) 125C) 60D) 15

Find |S|:Ex. 13a


* * *

2 * *

3 * *

4 * *

5 * *

1 * *

1 2 *

1 3 *

1 4 *

1 5 *

2 1 *

2 3 *

2 4 *

2 5 *

3 1 *

2 1 32 1 4

2 1 5

......

Hint: the probability tree is

A) 243B) 125C) 60D) 15

|S|, without replacement, order mattersEx. 13a

* * *

1 * *

1 2 *1 2 31 2 4

1 2 5

|E1| = 1|E2| = 3|E1| = 3

|E3| = 4|E2| = 4x3

|S| = 5|E3| = 5x4x3

|S| = 5x4x3x2x1

2x1

= n!

(n-k)!



Ex. 13b

Your picks: 12, 31, 10, 23, 8, 45

Draw result:a) if order matters.

b) if order does not matter:

|S| = n!

(n-k)!

|S| = ?

|S|, without replacement, order does not matterEx. 13b

2 *1 2 31 2 4

1 2 5

- Group the leaves that are equivalent when order is ignored

- How many in each group?

Group 1{1, 2, 3}

Group 2{1, 2, 4}2 *

2 1 32 1 4

2 1 5

......

6 in each group

more generally, k!

...


|S|, without replacement, order does not matterEx. 13b

2 *1 2 31 2 4

1 2 5Group 1{1, 2, 3}

Group 2{1, 2, 4}2 *

2 1 32 1 4

2 1 5

......

...


# leaves= n!

(n-k)!# leaves per

group = k!

Therefore, # groups =

n!

k!(n-k)!

DIY: important related problem / exercise

• You have a DNA sequenceACATG

• how many distinct genomes can you form with these 5 nucleotides (letters)

• answer is not 5!, because we have 2 A’s

• it is 5!/2 = 5! / (2! 1! 1! 1!)

• What is the general formula for a genome with nA A’s, nC C’s, nG G’s and nT T’s?

• Idea is similar to what we just saw (overcounting and dividing)


Ex. 13b

Your picks: 12, 31, 10, 23, 8, 45

Draw result:a) if order matters.


|S| = n!

(n-k)!

|S| = n!

k!(n-k)!


Ex. 13b

a) if order matters.


|S| = n!

(n-k)!

|S| = n!

k!(n-k)!

Notation:

n!

k!(n-k)!nk

Note:List where- order does not matter - items appear at most once

Set

=( )

‘Elementary permutations and combinations’

Def 6

a) if order matters.


|S| = n!

(n-k)!

|S| = n!

k!(n-k)!

‘Number of permutations’ Notation: nPk

# of list of size k, where each object is taken without replacement from n

possible objects’

‘Number of combinations’ Notation: nCk

# of sets of size k, where each object is taken without replacement from n

possible objects’

STAT 302 Intro to Probabilitybouchard/courses/stat302-sp2017-18//files/... · Prop. 3 De Morgan’s...

Documents

Transcript of STAT 302 Intro to Probabilitybouchard/courses/stat302-sp2017-18//files/... · Prop. 3 De Morgan’s...