Probability

Question 1

(S,P)

Answer 1

The sample space S of all elementary events x E S ; also called outcomes of experiment
A probability distribution P assigns a real number P(x) to every elementary event x E S such that ; Should all add up to one AND should not exceed one or go below zero

Question 2

Event

Answer 2

Event is a subset of the sample space S -> probability is sum of the probabilities of each one.

Question 3

Probability Distribution - uniform

Answer 3

A probability distribution is uniform if all outcomes are equally likely -> P(x) = 1/|S| to represent this (cardinality).

Question 4

Heads and Tails:

Answer 4

S = {H, T}
P(H) = P(T) = 1/2

S = {HH, HT, TH, TT}
P(HH) = P(HT) = P(TH) = P(TT) = 1/4

Question 5

Dice: (roll n amount of times)

Answer 5

S = {1,2,3,4,5,6}
P(S) = for every x in S : P(x) = 1/6

S = {1,…,6} -> {1,…,6}^n
P(x) = 1/(6^2), since S has 6^n elements

Question 6

P(E) (dice):

Answer 6

Number of sequences containing 6 / 6^3 if E was rolling at least one 6 in three turns.

Question 7

¬E = P without E

Answer 7

Then P(¬E) = 1 = P(E)

Question 8

Probability that at least one bit in a randomly generated sequence of 10 bits is 0.

Answer 8

If x exists in S, P(x) = (1/2)^10 = 1/(2^10) then
P(E) = 1 - 1/(2^10)

Question 9

P(R u C)

Answer 9

= P(R) + P(C) - P(R n C)
= 8/30 + 9/30 - 2/30 = 1/2

Question 10

P(A u B u C)

Answer 10

-> P(A) + P(B) + P(C)
- P(A n B) - P(A n C) - P(B n C)
+ P(A n B n C)

Question 11

S’ is a subset of S

Answer 11

S contains BB, GB, BG and GG, but let’s say it contains only 3 values, BB, GB, and BG
GG is not in S’
Then to find out the probability of BB, you do the probability of BB / probability of S’

Question 12

Probability that it is red considering it is circular

Answer 12

-> P(R|C)
P(R|C) -> P(R n C) / P(C) = 2/30 divided by 9/30 = 2/30 * 30/9 = 2/9

Question 13

P = {Red | Red, Green}

Answer 13

Not independent because it is certified that red exists (red = 1) because of both.

Question 14

Conditional Probability

Answer 14

The probability of an event occurring considering another event has already occurred.
P(A|B) for example.
You can then write P(A|B) = P(A n B) / P(B) since we do not need to include B due to it already happening.

Question 15

Rewrite P(A|B) = P(A n B) / P(B)

Answer 15

P(A n B) = P(A|B) * P(B)
P(B) = P(A n B) / P(A|B)

Question 16

Probability of getting a king then a queen from a deck?
Multiplication Rule

Answer 16

P(K n Q) = P(K) * P(Q|K) = 0.6%

Question 17

Independent

Answer 17

P(A n B) = P(A) * P(B)
Neither effect each other.
If we apply this to the king and queen question, it would result in 4/52 * 4/52, not 4/51.

P(F,G) = P(F) * P(G) since they do not directly effect each other (like choosing a queen then a king would do from a pack of cards ; toothache and weather is another example).

Question 18

Bayes’ Theorem

Answer 18

If P(A) > 0 then:
P(B|A) = P(A|B) * P(B) / P(A)

Rewritten as:
P(B|A) = P(A|B) * P(B) / P (A|B ) ×P (B ) + P (A|¬B ) ×P (¬B )
Denominator = P(A|B^2) + P(A|¬B^2)
B’s cancel out
P(A) + P(A) = P(A)

Question 19

Random Variable

Answer 19

The result from a random experiment.
S = {car,train,dog,cat}
F(car) = 1, F(train) = 1, F(dog) = 2, F(cat) = 2
(F = 1) denotes {s exists in S | F(s) = 1} = {car, train}

Question 20

S = {1,2,3,4,5,6}^2 = for two dice rolls
What is the probability for a and b to equal 12?

Answer 20

P(ab) = 1/36 for every ab which exists in S
F(ab) = a + b
r = 12 (arbitrarily chosen)
P(F = r) = P(ab | F(a|b) = r)

Question 21

¬(F = r)

Answer 21

{s | F(s) != r}
P(Die != 1) = P¬(Die = 1) = 5/6

Question 22

Probability Distribution For Random Variable

Answer 22

Gives the probabilities of all the possible values of the random variable

Question 23

Joint Probability Distribution

Answer 23

The probability distribution for two things combined, like the chances of a toothache and if the weather is sunny.

Question 24

Bold P

Answer 24

Conditional Distribution

Question 25

Conditional Distribution

Answer 25

P(F|G) = P(F = r| G = s)
Using Product Rule
P(F = r1,G = s1) = P(F = r1|G = s1)P(G=s1)
P(F = r2,G = s2) = P(F = r2|G = s2)P(G=s2)

Question 26

Probability Inference

Answer 26

P(Q|E1 = e1,…,En = en)
Conditional all the way to n.
Way of saying conditional queries can be joint.

Question 27

Probability Inference Example
P(Cavity|Toothache=1)

Answer 27

P(Cavity = 1|Toochache = 1)
P(Cavity = 0|Toochache = 1)
P(Cavity = 1|Toochache = 1) = P(Cavity = 1|Toochache = 1) / P(Toothache = 1) = 0.12/0.2
P(Cavity = 0|Toochache = 1) = P(Cavity = 0|Toochache = 1) / P(Toothache = 1) = 0.08 / 0.2
0.2 = normalization constant
(0,6,0.4)

Question 28

Conditional Independence

Answer 28

G, F are conditionally independent if
P(G, F|H1,…,Hn) = P(G|H1,…,Hn) * P(F|H1,…,Hn)

Let’s say we have two independent variables, weather and dentistry variables.
P (Weather = sunny, Toothache = r1, Catch = r2, Cavity = r3)
= P (Weather = sunny)P (Toothache = r1,Catch = r2,Cavity = r3)

Question 29

Belief Networks

Answer 29

Way of representing independence.

Question 30

Full Joint Probability Distribution with BN

Answer 30

P(Background, Works_Hard, Answers, Grade) -> P(Background) * P(Works_Hard) * P(Answers|Background|Works_Hard) * P(Grades|Answers)

Question 31

Expected Value

Answer 31

The value you are expected to get back depending on random variables.

Question 32

Suppose you roll a fair die, what is the expected value?

Answer 32

Random Variable F with values {1,2,3,4,5,6}
P(F=x) = 1/6 for all x in {1,2,3,4,5,6}
E(F) = 1 * P(F = 1) + 2 * P(F = 2) … = 3(1/2) = 3.5

Question 33

Distributive Law

Answer 33

2 (3 + 1) = (2 * 3) + (2 * 1)