Probability 2

Area of a Square

100%

Area of Green Square (X)

X = 25%

Area of NOT-X

~X = 75%

Formula

Area(~X) = 1 – Area(X)

Area of X

X = 25%

Area of A

X = 25%

A = 25%

Area of X v A

X = 25%

A = 25%

50%

Formula

If X and A are non-overlapping, then Area(X v A) = Area(X) + Area(A)

Area of Y

Y = 50%

Area of Z

Z = 50%

Area of Y or Z

Y v Z = 75%

Formula

Area(Y v Z) = Area(Y) + Area(Z) – Area(Y & Z)

Area of Y & Z

Y & Z = 25%

Independence

Y and Z are independent: knowing that a point is in Y does not increase the probability that it’s in Z, because half of the points in Y are in Z and half are not.

Formula

If Y and Z are independent, thenArea(Y & Z) = Area(Y) x Area(Z)

Area of Z

Z = 50%

Area of B

50%

Area of B v Z

62.5%

Area(Z & B)

Z & B = 37.5%

Correlated Areas

B and Z are not independent. 75% of the points in Z are also in B. If you know that a point is in Z, then it is a good guess that it’s in B too.

Formula

Area(B & Z) = Area(B) x Area(Z/ B)

This is the percentage of B that is in Z: 75%

Area(Z & B)

Z & B = 37.5%

Area(Z & B)

Z & B = 37.5% Z = 50%

Conditional Areas

Area(Z/ B) = Area(Z & B) ÷ Area (B)= 37.5% ÷ 50%

= 75%

Formula

Area(B & Z) = Area(B) x Area(Z/ B)= 50% x 75%

= 37.5%

Area(Z & B)

Z & B = 37.5%

Rules

Area(~P) = 1 – Area(P)Area(P v Q) = Area(P) + Area(Q) – Area(P & Q)

Area(P v Q) = Area(P) + Area(Q) for non-overlapping P and QArea(P & Q) = Area(P) x Area(Q/ P)

Area(P & Q) = Area(P) x Area(Q) for independent P and Q

Rules

Pr(~φ) = 1 – Pr(φ)Pr(φ v ψ) = Pr(φ) + Pr(ψ) – Pr(φ & ψ)

Pr(φ v ψ) = Pr(φ) + Pr(ψ) when φ and ψ are mutually exclusivePr(φ & ψ) = Pr(φ) x Pr(ψ/ φ)

Pr(φ & ψ) = Pr(φ) x Pr(ψ) when φ and ψ are independent

In-Class Exercises

1. Pr(P) = 1/2, Pr(Q) = 1/2, Pr(P & Q) = 1/8, what is Pr(P v Q)?

2. Pr(R) = 1/2, Pr(S) = 1/4, Pr(R v S) = 3/4, what is Pr(R & S)?

3. Pr(U) = 1/2, Pr(T) = 3/4, Pr(U & ~T) = 1/8, what is Pr(U v ~T)?

Known: Pr(P) = 1/2, Known: Pr(Q) = 1/2, Known: Pr(P & Q) = 1/8

Unknown: Pr(P v Q)

Rules




Pr(P v Q)

Pr(φ v ψ) = Pr(φ) + Pr(ψ) – Pr(φ & ψ)

Pr(P v Q) = Pr(P) + Pr(Q) – Pr(P & Q) = 1/2 + Pr(Q) – Pr(P & Q) = 1/2 + 1/2 – Pr(P & Q) = 1/2 + 1/2 – 1/8 = 7/8

Known: Pr(P) = 1/2, Known: Pr(Q) = 1/2, Known: Pr(P & Q) = 1/8

Known: Pr(R) = 1/2 Known: Pr(S) = 1/4 Known: Pr(R v S) = 3/4

Unknown: Pr(R & S)?

Not Helpful: More than One Unknown




This Is What You Want




Pr(R & S)


Pr(R v S) = Pr(R) + Pr(S) – Pr(R & S)3/4 = Pr(R) + Pr(S) – Pr(R & S)3/4 = 1/2 + Pr(S) – Pr(R & S)3/4 = 1/2 + 1/4 – Pr(R & S)3/4 = 3/4 – Pr(R & S)

Known: Pr(R) = 1/2 Known: Pr(S) = 1/4 Known: Pr(R v S) = 3/4

Known: Pr(U) = 1/2Known: Pr(T) = 3/4 Known: Pr(U & ~T) = 1/8

Unknown: Pr(U v ~T)?

Rules




Pr(~T)

Pr(~φ) = 1 – Pr(φ)

Pr(~T) = 1 – Pr(T) = 1 – 3/4 = 1/4

Known: Pr(U) = 1/2Known: Pr(T) = 3/4 Known: Pr(U & ~T) = 1/8Known: Pr(~T) = 1/4

Pr(U v ψ) = Pr(U) + Pr(ψ) – Pr(U & ψ)


Pr(U v ~T) = Pr(U) + Pr(~T) – Pr(U & ~T)


Pr(U v ~T)

Pr(U v ~T) = Pr(U) + Pr(~T) – Pr(U & ~T) = 1/2 + Pr(~T) – Pr(U & ~T) = 1/2 + 1/4 – Pr(U & ~T) = 1/2 + 1/4 – 1/8 = 5/8


More Exercises

4. Suppose I flip a fair coin three times in a row. What is the probability that it lands heads all three times?

5. Suppose I flip a fair coin four times in a row. What is the probability that it does not land heads on any of the flips?

Problem #4

4. Suppose I flip a fair coin three times in a row. What is the probability that it lands heads all three times?

Known: Pr(F) = 1/2Known: Pr(S) = 1/2Known: Pr(T) = 1/2

Unknown: Pr((F & S) & T)

Rules




Pr((F & S) & T)

Pr(φ & ψ) = Pr(φ) x Pr(ψ)

Pr((F & S) & T) = Pr(F & S) x Pr(T) = Pr(F) x Pr(S) x Pr(T) = 1/2 x 1/2 x 1/2 = 1/8

Problem #5

5. Suppose I flip a fair coin four times in a row. What is the probability that it does not land heads on any of the flips?Known: Pr(F) = 1/2Known: Pr(S) = 1/2Known: Pr(T) = 1/2Known: Pr(L) = 1/2

Unknown: Pr((~F & ~S) & (~T & ~L))

Rules




Using ~ Rule

Pr(~F) = 1 – 1/2 = 1/2Pr(~S) = 1 – 1/2 = 1/2Pr(~T) = 1 – 1/2 = 1/2Pr(~L) = 1 – 1/2 = 1/2

Pr((~F & ~S) & (~T & ~F))

Pr(φ & ψ) = Pr(φ) x Pr(ψ)

Pr((~F & ~S) & (~T & ~F)) = Pr(~F & ~S) x Pr(~T & ~F) = Pr(~F) x Pr(~S) x Pr(~T & ~F) = Pr(~F) x Pr(~S) x Pr(~T) x Pr(~F) = 1/2 x 1/2 x 1/2 x 1/2 = 1/16

Conditional Probability

Area of P

P = 50%

Area of Q

Q = 50%

Area(P & Q)

P & Q = 37.5%

Area(P & Q) out of Area(Q)

P & Q = 75%

Conditional Areas

Area(P/ Q) means: the percentage of (P & Q) points out of all Q-points.

Area(P/ Q) = Area(P & Q)

Area(Q)

Conditional Probabilities

Pr(P/ Q) means: the percentage of (P & Q) possibilities out of all Q-possibilities.

Pr(P/ Q) = Pr(P & Q)

Pr(Q)

Probabilistic Generalizations

Our probabilistic generalizations usually express conditional probabilities:90% of bankers are rich ≠ the probability of someone being rich is 90%≠ the probability of someone being a banker is 90%≠ the probability of someone being a rich banker is 90%= the probability of someone being rich assuming that they are a banker is 90%

Coin Flips

Suppose I flip a coin twice.

The probability that it will land heads on the first flip is 50%.

The probability that it will land heads on the second flip is 50%.

Coin Flips

First SecondHeads HeadsHeads TailsTails HeadsTails Tails

Coin Flips

Assuming nothing, what is the probability that both coins land heads?

Pr(F & S) = ?

Pr(F & S) = 1/4


How Did We Calculate That?

Since two coin flips are independent, we know:

Pr(F v S) = Pr(F) x Pr(S) = 50% x 50% = 25%

Coin Flips

Assuming that one of the coin flips lands heads, what is the probability that the other one also lands heads?

Pr(F & S/ F v S) = ?

Ignore the Possibilities with No Heads


Pr(F & S/ F v S) = 1/3

First SecondHeads HeadsHeads TailsTails Heads


Pr(F & S/ F v S) = = = = 1/3Pr((F & S) & (F v S))Pr(F v S)

Pr(F& S)Pr(F v S)

25%75%

Coin Flips

Assuming that the first coin flip lands heads, what is the probability that the other one also lands heads?

Pr(F & S/ F) = ?

Ignore Possibilities Where First Is Not Heads


Pr(F & S/ F) = 50%

First SecondHeads HeadsHeads Tails


Pr(F & S/ F) = = = = 1/2Pr((F & S) & F)Pr(F)

Pr(F& S)Pr(F)

25%50%

Bayes’ Theorem

Simple Algebra

Pr(B/ A) = Pr(B & A) ÷ Pr(A)Pr(A) x Pr(B/ A) = [Pr(A) x Pr(B & A)] ÷ Pr(A)Pr(A) x Pr(B/ A) = [Pr(A) x Pr(B & A)] ÷ Pr(A)Pr(A) x Pr(B/ A) = Pr(B & A)Pr(A & B) = Pr(B/ A) x Pr(A)

Bayes’ Theorem

Pr(A/ B) = Pr(A & B) ÷ Pr(B)Pr(A/ B) = [Pr(B/ A) x Pr(A)] ÷ Pr(B)

Bayes’ Theorem

Baye’s theorem lets us calculate the probability of A conditional on B when we have the probability of B conditional on A.

Base Rate Fallacy

• There are ½ million people in Russia are affected by HIV/ AIDS.

• There are 150 million people in Russia.

Base Rate Fallacy

Imagine that the government decides this is bad and that they should test everyone for HIV/ AIDS.

The Test

If someone has HIV/ AIDS, then :• 95% of the time the test will be

positive (correct)• 5% of the time will it be negative

(incorrect)

The Test

If someone does not have HIV/ AIDS, then:• 95% of the time the test will be

negative (correct)• 5% of the time will it be positive

(incorrect)

Suppose you test positive. We’re interested in the conditional probability: what is the probability you have HIV assuming that you test positive.

We’re interested in Pr(HIV = yes/ test = pos)

Known: Pr(sick) = 1/300Known: Pr(positive/ sick) = 95%Known: Pr(positive/ not-sick) = 5%

Unknown: Pr(positive)Unknown: Pr(sick/ positive)

Pr(positive)

Pr(positive) = True positives + false positives= [Pr(positive/ sick) x Pr(sick)] + [Pr(positive/ not-sick) x Pr(not-sick)]= [95% x 1/300] + [5% x 299/300]= 5.3%

Known: Pr(sick) = 1/300Known: Pr(positive/ sick) = 95%Known: Pr(positive/ not-sick) = 5%

Known: Pr(sick) = 1/300Known: Pr(positive/ sick) = 95%Known: Pr(positive) = 5.3%

Unknown: Pr(sick/ positive)

Pr(sick/ positive)

Pr(A/ B) = [Pr(B/ A) x Pr(A)] ÷ Pr(B)

Pr(sick/ positive) = [Pr(positive/ sick) x Pr(sick)] ÷ Pr(positive) = [95% x Pr(sick)] ÷ Pr(positive) = [95% x 1/300] ÷ Pr(positive) = [95% x 1/300] ÷ 5.3% = 5.975% Known: Pr(sick) = 1/300

Known: Pr(positive/ sick) = 95%Known: Pr(positive) = 5.3%

Probability 2

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