Stat Theory: Chapter 1 &2: Probability

Front 1

Proof: If A⊂B and B⊂C, imply A⊂C

Back 1

Suppose x ∈ A.

Since A⊂B, x ∈ B.

Since B⊂C, x ∈ C

∴ A⊂C

Front 2

DeMorgan's Law 1:

(A∩B)' = ?

Back 2

A'UB'

Front 3

DeMorgan's Law 2:

(AUB)' = ?

Back 3

A'∩B'

Front 4

Multiplicative Law: The probability of the INTERSECTION of two events A and B.

P(A∩B) = ?

Back 4

P(A) * P(B|A)

and

P(B) * P(A|B)

Front 5

Multiplicative Law for 3 Events

P(A∩B∩C) = ?

Back 5

P[(A∩B) U C]

=

P(A∩B) * P(C|A∩B)

=

P(A) * P(B|A) * P(C|A∩B)

Front 6

Additive Law : The probability of the UNION of two events A and B.

P(AUB) = ?

Back 6

P(A) + P(B) - P(A∩B)

Front 7

If P(A)>0 and P(B)>0, then mutually exclusive implies...

Back 7

dependence.

Front 8

If P(A∩B)=0, then ...

Back 8

... A and B are mutually exclusive.

Front 9

P(A|B) ⇔ (formula)

Back 9

⇔ P(A∩B) / P(B)

Front 10

P(B|A) ⇔ (formula)

Back 10

⇔ P(B∩A) / P(A)

Front 11

Mutually Exclusive ⇒

Back 11

Dependence

(But dependence ≠ M.E.)

Front 12

Independence ⇒

Back 12

Not Mutually Exclusive

(But Not M.E. ≠ Independence)

Front 13

Law of Total Probability

P(A) = formula

Back 13

P(A) = ∑ P(A|Bi) * P(Bi) for i=1 to k.

(Conditions:

Let B1, B2, ... , Bk be a partition of the sample space Si.

B1∪B2∪...∪Bk = Si and Bi∩Bj = ∅ for i≠j)

Front 14

Baye's Rule

P(Bj|A) = formula

Back 14

P(Bj|A) = [ P(Bj)*P(A|Bj)] / [∑(P(Bi)*P(A|Bi)]

where Bj is constant

and i=1 to n

Front 15

P(A|AUB) = (equivalent formula)

Back 15

P(A) / P(AUB)

Front 16

P(A|A∩B) = (equivalent formula)

Back 16

P(A∩B) / P(A∩B) = 1

This is because we already know that event A has occurred.

Front 17

P(A∩B|AUB) = (equivalent formula)

Back 17

P(A∩B) / P(AUB)