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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.
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.
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
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)?
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
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
Not Helpful: More than One Unknown
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
This Is What You Want
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
Pr(R & S)
Pr(φ v ψ) = Pr(φ) + Pr(ψ) – Pr(φ & ψ)
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
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
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(φ v ψ) = Pr(φ) + Pr(ψ) – Pr(φ & ψ)
Known: Pr(U) = 1/2Known: Pr(T) = 3/4 Known: Pr(U & ~T) = 1/8Known: Pr(~T) = 1/4
Pr(φ v ψ) = Pr(φ) + Pr(ψ) – Pr(φ & ψ)
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)
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)
Known: Pr(U) = 1/2Known: Pr(T) = 3/4 Known: Pr(U & ~T) = 1/8Known: Pr(~T) = 1/4
Pr(U v ~T) = Pr(U) + Pr(~T) – Pr(U & ~T)
Known: Pr(U) = 1/2Known: Pr(T) = 3/4 Known: Pr(U & ~T) = 1/8Known: Pr(~T) = 1/4
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
Known: Pr(U) = 1/2Known: Pr(T) = 3/4 Known: Pr(U & ~T) = 1/8Known: Pr(~T) = 1/4
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(~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
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
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
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 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%.
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) = ?
How Did We Calculate That?
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
First SecondHeads HeadsHeads TailsTails HeadsTails Tails
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
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%
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