Lecture 2: Probability - Oxford Statistics › ... › lecture2.pdfLecture 2: Probability 21st of October 2015 24 / 56. Probability of independent events In probability theory we say

Lecture 2: Probability

21st of October 2015

Lecture 2: Probability 21st of October 2015 1 / 56

Outline

In this and the following lecture we will learn about

why we need to learn about probability

what probability is

how to assign probabilities

how to manipulate probabilities and calculate probabilities of complexevents


populationabout aHypothesis

STATISTICSStudyDesign

Propose anexperiment

Takea

sample

STATISTICALTEST

ExamineResults


Why do we need probability?

The conclusion we make about a hypothesis depend upon the samplewe take.

The sample we take may lead us to the wrong conclusion about thepopulation.

We need to know what the chances are of this happening.

Probability is the study of chance.


Baby boom example

Hypothesis: Boys weight more than girls at birth.

Data from our sample:Sample mean of boys weights = xboys = 3375.308Sample mean of girls weights = xgirls = 3132.444Difference between sample means: D = xboys − xgirls = 242.8632

Statistical analysis:

Does this allow us to conclude that in the population boys are bornheavier than girls?

Is the difference higher than would be expected by chance?

If yes then we can be confident that in the population boys areheavier than girls on average at birth.


Chance experiment

In probability we consider chance experiment: situation whose outcome isuncertain.

Examples:

Rolling a dice

Tossing a coin

Counting number of people with glasses in the class

Occurence of a given phenotype in a population

Measuring glucose

Drug trial

Launching a rover on Mars


Sample space

The sample space for an experiment is the set of all possible outcomes.

Examples:

Rolling a dice: Ω = 1, 2, 3, 4, 5, 6Tossing a coin: Ω = Head, TailCounting people in the class: Ω = 0, 1, 2 . . . nOccurence of a given phenotype in a population (percentage):Ω = [0, 100]

Drug trial : Measure of disease progress

Launching a rover on Mars: Ω = Crash, Land


Puzzle 1

You are dealt two cards successively without replacement from a standarddeck of 52 playing cards.

Find the probability that the first card is a six and the second card is aQueen.


Puzzle 2

You have reached the final of a game show. The host shows you threedoors and tell you that there is a prize behind one of the door. You pick adoor. The host then opens one of the door you did not pick: there is noprice behind this door. He then asks you if you want to change from thedoor you chose to the other remaining door.

Is it to your advantage to switch your choice?


Simple example: Rolling a die

Consider the experiment of rolling a fair six-sided die.

The sample space of our experiment is the set S = 1, 2, 3, 4, 5, 6.

A particular outcome is called a sample point.Example: The outcome ”the top face shows a three” is the sample point 3.

A collection of possible outcomes is called an event.Examples:

The event A1, that the die shows an even number is the subsetA1 = 2, 4, 6 of the sample space.

The event A2 that the die shows a number larger than 4 is the subsetA2 = 5, 6.


Calculating simple probabilities

The probability of an event is a number between 0 and 1, inclusive, thatindicates how likely the event is to occur.

The higher the probability of an event, the more certain we are that theevent will occur.

In some settings (like the example of the fair die considered above) it isnatural to assume that all the sample points are equally likely.

In this case, we can calculate the probability of an event A as

P (A) =|A||S|

,

where |A| denotes the number of sample points in the event A.


Die example (continued)

S = 1, 2, 3, 4, 5, 6 A1 = 2, 4, 6 A2 = 5, 6

P (A1) =|A1||S|

=3

6=

1

2

S

2

1 3

4 6

5

A1

P (A2) =|A2||S|

=2

6=

1

3

S

2

1 3

4 6

5 A2


Intersection

What about P (face is even, and larger than 4) ?

We can write this event in set notation as A1 ∩A2.

This is the intersection of the two events, A1 and A2

i.e the set of elements which belong to both A1 and A2.

A1 ∩A2 = 6 ⇒ P (A1 ∩A2) =|A1 ∩A2||S|

=1

6

S

2

1 3

4 6

5 A2

A1


Union

What about P (face is even, or larger than 4) ?

We can write this event in set notation as A1 ∪A2.

This is the union of the two events, A1 and A2

i.e the set of elements which belong either A1 and A2 or both.

A1 ∪A2 = 2, 4, 5, 6 ⇒ P (A1 ∪A2) =|A1 ∪A2||S|

=4

6=

2

3

S

2

1 3

4 6

5 A2

A1


Complement

What about P (face is not even) ?

We can write this event in set notation as Ac1.

This is the complement of the event, A1

i.e the set of elements which do not belong to A1.

Ac1 = 1, 3, 5 ⇒ P (Ac

1) =|Ac

1||S|

=3

6=

1

2

S

2

1 3

4 6

5

A1


Probability in more general settings

In many settings, either the sample space is infinite or all possibleoutcomes of the experiment are not equally likely. We still wish toassociate probabilities with events of interest.

Luckily, there are some rules/laws that allow us to calculate andmanipulate such probabilities with ease.


Probability Axioms (Building Blocks)

There are three axioms which we need in order to develop our laws

1 0 ≤ P (A) ≤ 1 for any event A.

2 P (S) = 1.

3 If A1, . . . , An are mutually exclusive events, then

P (A1 ∪ . . . ∪An) = P (A1) + . . . + P (An).

A set of events are mutually exclusive if at most one of the eventscan occur in a given experiment.


Complement Law

If A is an event, the set of all outcomes that are not in A is called thecomplement of the event A, denoted A.

The rule isA = 1− P (A)

Example:S = the set of students at Oxford. We are picking a student at random.A = the event that the randomly selected student suffers from depression

We are told that 8% of students suffer from depression, so P (A) = 0.08.What is the probability that a student does not suffer from depression?

P (A) = 1− 0.08 = 0.92.


Addition Law (Union)

Suppose,

A = a randomly selected student from the class has brown eyesB = a randomly selected student from the class has blue eyes

What is the probability that a student has brown eyes OR blue eyes?

This is the union of the two events A and B, denoted A∪B.

In general for two events

P (A ∪B) = P (A) + P (B)− P (A ∩B)


A B

A∩B

S

If we simply add P (A) and P (B) we will count the sample points in theintersection A ∩B twice and thus we need to subtract P (A ∩B) fromP (A) + P (B) to calculate P (A ∪B).


Example: SNPs

Single nucleotide polymorphisms (SNPs) are nucleotide positions in agenome which exhibit variation amongst individuals in a species. In somestudies in humans, SNPs are discovered in European populations.

Suppose that of such SNPs,

70% also show variation in an African population,

80% show variation in an Asian population

and 60% show variation in both the African and Asian population.

Suppose one such SNP is chosen at random, what is the probability that itis variable in either the African or the Asian population?


Write A for the event that the SNP is variable in Africa, and B for theevent that it is variable in Asia.

We are told

P (A) = 0.7

P (B) = 0.8

P (A ∩B) = 0.6.

We require P (A ∪B). From the addition rule:

P (A ∪B) = P (A) + P (B)− P (A ∩B)

= 0.7 + 0.8− 0.6

= 0.9.


Frequency concept of probability

In tossing a fair coin, both outcomes have the same probability.

This is not the case in general.

Example: Drug effect.

If the drug is found to be effective 30% of the time, we might assigna probability of 0.3 that the probability is effective and a probabilityof 0.7 that it is not effective.

This illustrate the frequency concept of probability:if we have a probability p that an experiment will result in outcomeA, then if we repeat this experiment a large number of times weshould expect that the fraction of times that A will occur is about p.


Independent events

Two events are independent if the occurrence of one of the eventsgives us no information about whether or not the other event occurs.

Independent events have no influence on each other

Example: You toss a fair coin and it comes up ”Head” three times, what isthe chance that the next toss will also be a ”Head”?

The chance is simply just like ANY toss of the coin.

What happened in the past will not affect the current toss!

The coin tosses are said to be independent.


Probability of independent events

In probability theory we say that two events, A and B, are independent ifand only if the probability that they both occur is equal to the product ofthe probabilities of the two individual events, i.e.

P (A ∩B) = P (A)P (B)


Example: Blue eyes and blond hair

Blond hair and blue eyes are two traits often thought to be related toone another.

Calculate the probability of having both blond hair and blue eyestogether if 20% of people have blond hair, 30% of people have blueeyes, and blond hair and blue eyes are independent.

Solution:

A = blond hair

B = blue eyes

Aussming independence of A and B

P (A ∩B) = P (A)P (B) =20

100× 30

100=

600

10000=

6

100


Example: Blue eyes and blond hair

Suppose that as before, 20% of people have blond hair, and 30% ofpeople have blue eyes.

If 15% of people have both blond hair and blue eyes, are blond hairand blue eyes independent?

Solution:

If blue eyes and blond hair are independent events, the probability ofhaving both blond hair and blue eyes is:

P (A ∩B) = P (A)P (B) =6

100

Since 15% of people have both blond hair and blue eyes and15100 6=

6100 then blue eyes and blond hair are not independent events.


Summary of this lecture

Define chance experiment, sample space and event

Understand the concept of the probability of an event

Show how to manipulate intersection, union and complement ofevents

Show how to compute analytically probability for simple chanceexperiments

Define independence between event


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Lecture 2: Probability - Oxford Statistics › ... › lecture2.pdfLecture 2: Probability 21st of October 2015 24 / 56. Probability of independent events In probability theory we say