Probability (Equations)

Question 1

P(R u C)

Answer 1

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

Question 2

P(A u B u C)

Answer 2

-> 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 3

Probability that it is red considering it is circular

Answer 3

-> 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 4

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

Answer 4

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

Question 5

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

Answer 5

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

Question 6

Independent

Answer 6

P(F,G) = P(F) * P(G)
P(A n B) = P(A) * P(B)
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).
Don’t use the answer from the equation as the answer. We use this equation to determine if it IS independent, since the equation would hold up correctly.

Question 7

Bayes’ Theorem

Answer 7

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

Question 8

Bayes’ Theorem Alternative

Answer 8

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 9

Random Variable
S = {car,train,dog,cat}
F(car) = 1, F(train) = 1, F(dog) = 2, F(cat) = 2
What is F=1?

Answer 9

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 10

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

Answer 10

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)
P (F = 12) = P ({ab |F (ab) = 12}) = P (66) = 1/36

Question 11

¬(F = r)

Answer 11

{s | F(s) != r}

Question 12

Conditional Distribution

Answer 12

P(F|G) = P(F = r| G = s)
Using Multiplication 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 13

Probability Inference

Answer 13

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

Question 14

Probability Inference Example
P(Cavity|Toothache=1)

Answer 14

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 15

Conditional Independence

Answer 15

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 16

Normalisation
S’ is a subset of S
4 coins, 2 low value, 2 high.
What is P( 3 tails | at least one low value is tails) if they are all flipped once.

Answer 16

P of one low value = {HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT} = 12/16, assuming the first two were low value
P of 3 tails = S’ = {HTTT, THTT, TTHT, TTTH, TTTT} = 5/16
Normalisation = 3 tails / S’ = 5/16 / 12/16 = 5/12
Normalisation Rule = P( A | B ) = P(A) / P(B)

Question 17

Is the event that at least three of the coins come up tails independent of the event that at least one of the low-value coins comes up tails?

Answer 17

Probability of 3 = S’ = 5/16
Probability of one LV = 12/16
Since S’ is a subset of LV, because LV contains all elements S’ has, then P(LV and S’) = P(S’) = 5/16, represented if they weren’t independent.
But S(LV and S’) = P(S’) * S(LV) if they were independent which is not 5/16.

Question 18

P(a) expanded

Answer 18

P(A | B)P(B) + P(A | ¬B)P(¬B)