Probability Flashcards
(S,P)
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
Event
Event is a subset of the sample space S -> probability is sum of the probabilities of each one.
Probability Distribution - uniform
A probability distribution is uniform if all outcomes are equally likely -> P(x) = 1/|S| to represent this (cardinality).
Heads and Tails:
S = {H, T}
P(H) = P(T) = 1/2
S = {HH, HT, TH, TT}
P(HH) = P(HT) = P(TH) = P(TT) = 1/4
Dice: (roll n amount of times)
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
P(E) (dice):
Number of sequences containing 6 / 6^3 if E was rolling at least one 6 in three turns.
¬E = P without E
Then P(¬E) = 1 = P(E)
Probability that at least one bit in a randomly generated sequence of 10 bits is 0.
If x exists in S, P(x) = (1/2)^10 = 1/(2^10) then
P(E) = 1 - 1/(2^10)
P(R u C)
= P(R) + P(C) - P(R n C)
= 8/30 + 9/30 - 2/30 = 1/2
P(A u B u C)
-> P(A) + P(B) + P(C)
- P(A n B) - P(A n C) - P(B n C)
+ P(A n B n C)
S’ is a subset of S
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’
Probability that it is red considering it is circular
-> P(R|C)
P(R|C) -> P(R n C) / P(C) = 2/30 divided by 9/30 = 2/30 * 30/9 = 2/9
P = {Red | Red, Green}
Not independent because it is certified that red exists (red = 1) because of both.
Conditional Probability
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.
Rewrite P(A|B) = P(A n B) / P(B)
P(A n B) = P(A|B) * P(B)
P(B) = P(A n B) / P(A|B)
Probability of getting a king then a queen from a deck?
Multiplication Rule
P(K n Q) = P(K) * P(Q|K) = 0.6%
Independent
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).
Bayes’ Theorem
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)
Random Variable
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}
S = {1,2,3,4,5,6}^2 = for two dice rolls
What is the probability for a and b to equal 12?
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)
¬(F = r)
{s | F(s) != r}
P(Die != 1) = P¬(Die = 1) = 5/6
Probability Distribution For Random Variable
Gives the probabilities of all the possible values of the random variable
Joint Probability Distribution
The probability distribution for two things combined, like the chances of a toothache and if the weather is sunny.
Bold P
Conditional Distribution
