Chapter 6: The Mathematics of Probability

Question 1

❮ i,j ❯

Answer 1

Ordered Pair (Probability Calculus)

i = outcome of first roll
j = outcome of second roll.

Question 2

Example: Ordered pairs ❮ 1,4 ❯ and ❮ 4,1 ❯ are __________ different ___________.

Answer 2

two; outcomes

Question 3

“S”

Answer 3

Represents all possible outcomes in a random experiment.
“Sample Space”

Question 4

“Sample Space” must be either _____________ or ____________ _____________.

Answer 4

finite; countably infinite

Question 5

Event A

Answer 5

Subset of the Sample Space (S).

Question 6

An Event can be ___ and ___, ____ or ___, or _____-___ and _____-___.

Answer 6

A and B; A or B; not-A and not-B

Question 7

Aᶜ

Answer 7

Complement of Event A

The complement of Event A is all other Events in the Sample Space.

Question 8

Sᶜ

Answer 8

Complement of Sample Space (S)

Sᶜ = ∅ (where “∅” is a null sign, meaning “empty set”).

The complement of a Sample Space (S) is an empty set.

Question 9

p

Answer 9

p = Probability Measure

Takes an Event as it’s argument, and assigns a real number 0-1 as the probability of that event.

Question 10

Probability Calculus (Set of 3 Axioms)

Answer 10

1) Probability of an Event is a real number between 0 and 1.

2) Probability of Sample Space (S) is 1.

3) If two Events are mutually exclusive (meaning both can’t happen simoultaneously), then the probability of one Event happening equals the probability of the first plus that of the second.

Question 11

Mutually exclusive

Answer 11

both events can’t happen at the same time.

Question 12

Probability Calculus (Formalized)

Answer 12

1) p(A) = 1≥p(A)≥0
2) p(S) = 1
3) If A ᑎ B = ∅, then p(A ᑌ B) = p(A) + p(B)

Question 13

A ᑎ B

Answer 13

“A intersection B”

Only elements that are in both A and B

When A ᑎ B = ∅, that means A and B don’t have any intersecting elements, hence they’re mutually exclusive

Question 14

A ᑌ B

Answer 14

“A union B”

Contains all elements by A or B, or both.

Question 15

When representing the Probability Calculus in _______________ ___________, A and B are no longer “___________,” they become “_________________.”

Answer 15

propositional logic; Events; Propositions

Question 16

If Propositions ____ and ___ are ______________-______________ (meaning both can’t occur at the same time such that ___ ᐱ ___ = ____, AND that both ___ and ___ don’t have any __________ in common), then only one proposition must be _________, and the other must be _________.

Answer 16

A; B; mutually exclusive; A; B; ∅; A; B; elements; true; false

Question 17

Probability Calculus (Propositional Logic)

Answer 17

1) 0 ≥ p(A) ≥ 1

2) If proposition A is logically true (such that it’s impossible for A to be false), then p(A)=1.

3) If A ᐱ B = ∅, then p(A ᐯ B) = p(A) + p(B)

Question 18

The difference between 1st Set of Probability Calculus and Propositional Logic set of Probability Calculus

Answer 18

Probability calculus (on it’s own) assigns probabilities to events, THINGS THAT EXIST IN THE EXTERNAL WORLD.

Probability Calculus, formalized in propositional logic, assings probabilities to LINGUISTIC ENTITIES, aka. propositions.

Question 19

Theorem 6.1: p(A) + p(￢A) = 1

(Give steps on how to reach this answer)

Answer 19

p(A) + p(￢A) = 1

• A ᐯ ￢A is Logical Truth.
• Axiom 2 tells us, “If proposition A is a logical truth, then p(A) = 1
• Therefore, p(A ᐯ ￢A) = 1
• A and ￢A are Mutually Exclusive.
• Axiom 3 tells us, “If A ᐱ B = ∅, then p(A ᐯ B) = p(A) + p(B).”
  -Therefore, p(A ᐯ ￢A) = p(A) + p(￢A)

Taken together, p(A) + p(￢A) = p(A ᐯ ￢A) = 1

Question 20

Theorem 6.2: If A and B are logically equivalent, then p(A) = p(B)

(Give steps on how to reach this answer)

Answer 20

If A and B are logically equivalent, meaning they the same TRUTH VALUE,
then A and ￢B are mutually exclusive.

• Axiom 3: If A ᐱ B = ∅, then p(A ᐯ B) = p(A) + p(B).
• So it follows, that p(A ᐯ ￢B) = p(A) + p(￢B)

The compound sentence A ᐯ ￢B is true for some logical reasons (it’s Logically True).

• Axiom 2: If A is Logically True, then p(A) =1
• So it follows, that p(A ᐯ ￢B) = 1

Taken together, if p(A ᐯ ￢B) = 1, and p(A ᐯ ￢B) = p(A) + p(￢B), then p(A) + p(￢B) = 1.

Were not done yet though! Now we need to refer back to Theorem 6.1: p(A) + (￢A) = 1.

Plugging this theorem in for our sentence ￢B, we get p(B) + p(￢B) = 1

p(A) + p(￢B) = 1
p(B) + p(￢B) = 1
In both of these equations, we yield the same value for ￢B (1), so it follows that p(A) = p(B)

Question 21

Theorem 6.3 i) p(A v B) = p(A) + p(B) - p(A ∧ B)

Theorem 6.3 ii) p(A → B) = p(￢A) + p(B) - p(￢A ∧ B)

Answer 21

Proof pt. i)
- p(A v B) and p(￢A ∧ B) v p(A ∧ ￢B) v p(A ∧ B) are logically equivalent.
- Great! Axiom 2: If proposition A is logically equivalent, then p(A)=1
- However, p(￢A ∧ B) and p(A ∧ ￢B) v p(A v B) are mutuallly exclusive.
- Okay then, Axiom 3: If A ∧ B = ∅, then p(A v B) = p(A) + p(B)
- p(A v B) = p( (p(￢A ∧ B) ) v (p(A ∧ ￢B) v p(A v B) ) = p(￢A ∧ B) + [p(A ∧ ￢B) v p(A v B)]
- *You can simplify the problem, the rightmost equations are equivalent with A.
- p(A v B) = p(￢A ∧ B) + p(A)
- *Now we need p(B) to complete equation at it’s stated in the 3rd axiom (p(A v B) = p(A) + p(B)), and we need to eliminate p(￢A ∧ B). Why don’t we try adding p(A ∧ B) to both sides!
- p(A v B) + p(A ∧ B) = p(￢A ∧ B) + p(A) + p(A ∧ B)
- *You can simplify the problem, ￢A ∧ B and A ∧ B are equivalent w/ B.
- p(A v B) + p(A ∧ B) = p(B) + p(A)
- p(A v B) = p(B) + p(A) - p(A ∧ B)

Proof pt. ii)
- Same as Proof pt. i). Just be sure to substitute A w/ ￢A such that…
p(￢A v B) = p(B) + p(￢A) - p(￢A ∧ B)

Question 22

Rule for Conjunctive Probability with Independent Conjuncts

Answer 22

p(A Ʌ B) = p(A) × p(B)

Question 23

In most decision problems, ______________ is _______________.

Answer 23

probability; conditional

Question 24

p(A|B) (First Definition of A|B )

Answer 24

p(A|B) = p(A ∧ B) / p(B)

Question 25

p(A|B) (Second Definiton of A|B)

Answer 25

A is indpendent of B iff. p(A) = p(A|B)

Question 26

Theorem 6.4 “Independence”

Answer 26

If A is independent of B, then the probability of p(A) = p(A|B) / p(B)

Question 27

Baye’s Theorem Definition

Answer 27

• Calculates Conditional Probabilities
• Shows how conditional probabilies are “reserved.”

Question 28

Theorem 6.5: Baye’s Theorem – “Inverse Probability Law”

Answer 28

p(B|A) = p(B) x p(A|B) / p(A)

Question 29

Theorem 6.6: Bayes Theorem

Answer 29

p(B|A) = p(B) x p(A|B) / [p(B) x p(A|B)] + [p(￢B) x p(A|￢B)]

Question 30

Prior Probability

Answer 30

The unconditional probabilty p(B) (in Bayes’ Theorem).

May very from case to case, and can’t be known until there’s a body of data necessary to calculate it.

Question 31

A _________ to use in problems invovling Bayes’ Theorem, is assigning real _____________ (___-___) to the __________________ (____________) _________________ [p(B)]. In doing this we can find, on the other side of the Bayes’ Theorem problem, the _____________ _______________.

[See Notes for Example]

Answer 31

trick; numbers; prior (unknown) probability; posterior probability.

Question 32

In a problem, if you’re given ______ conditional prior and ____ unconditional priors, you should use the ___________ version of Bayes’ Theorem. If you’re given ____ conditional priors and ______ unconditional prior, you should employ the ___________ _____________ version of Bayes’ Theorem.

Answer 32

1; 2; simple; 2; 1; expanded denominator

Question 33

If a problem asks you for the probability of an event, you should find the probability of ________ getting at least getting that event, such that we’renot finding ___, but __ - __. Provide an example…

Answer 33

NOT; p; 1-p

If you’re asked to find the probability of rolling a one, on a six-sided fair dice, after rolling 4 times, you should calculate the probability of NOT rolling a one, out of 4 tries: 1 - (5/6 * 5/6 * 5/6 * 5/6).

Question 34

REMEMBER: Use the ___-___ rule when there’s not a direct or _____________ way to get at probabilities. Were supposed to use this rule when we can identify ____________ cases very easily–more easily than we can identify ____________ cases.

Answer 34

1-p; convenient; negative; positive

ex. (You’re being asked after rolling a six-sided fair die twice, what’s the probability of the Sum of the two rolls is 4. The only combinations, out of 36 are ⟨1,3⟩ ⟨2,2⟩ ⟨3,1⟩
that you’ll get a sum of four. Therefore, the p(S) = 3/36 = 1/12 = 0.083. HERE, it was alot more convenient to find the positive cases. This is a case in which you absolutely should NOT use the 1 - p rule, because then you’d be finding the negative cases–33/36 negative cases to be exact–when it’s SO much easier to just find the positive cases.)