Independence of Events in Probability
Events are often classified as
Dependent Events:
General Examples
If you park vehicle in unauthorized place, you are more likely to get a parking ticket
You must buy a lottery ticket to have a chance of winning
Independent Events
General Events:
Winning a Card game and running out of stock at home
Taking an rapido ride and getting a gift from friend
Two Events A and B are considered independent if any of the following is true
P(A|B) = P(A)
P(B|A) = P(B)
P(A ∩ B) = P(A)*P(B)
If none of the above holds good we call it dependent
Example-1: Consider the following events
A: odd number is rolled
B: even number is rolled
C: {1,2}
Are A and B independent?
P(A) = 3/6 = 1/2
P(B) = 3/6 = 1/2
P(A|B) = P (A ∩ B)/ P(B)
P (A ∩ B) = 0
p(A|B) = 0
# Since none of the above cases are true
A and B are dependent events
P(A) = 1/2
P(C) = 2/6 = 1/3
P(A|C) = 1/2 = P(A)
Events are independent
Example-2: Three Car brands X,Y and Z are ranked by judges
A – Car X is ranked higher than Y
B – Car X is ranked highest
C – Car X is ranked second best
D – Car X is ranked worst
Sample Space
E1: XYZ E2: XZY E3: YXZ
E4: YZX E5: ZXY E6: ZYX
A = { E1, E2, E5 }
B = { E1, E2 }
C = { E3, E5 }
D = { E4, E6}
P(A) = 3/6 = 1/2
P(A|B) = P(A ∩ B) / P(B) = 2/6 / 2/6 = 1
P(A|C) = P(A ∩ C)/P(C) = 1/6/2/6 = 1/2
P(A|D) = P(A ∩ D)/P(D) = 0
Result : A and C are independent & A,B and A, D are dependent
Multiplicative law of Probability
The probability of intersection of two events A and B is
P(A ∩ B )= P(B)*P(A|B)
If the events are independent “`P(A ∩ B) = P(A) * P(B)
Example: Lets consider the Dependent events, when you roll a die
A: Odd Number
B: number < 4
P(A) = 3/6 = 1/2
P(B) = 3/6 = 1/2
P(A|B)= 2/3
P(B|A) = 2/3
P(A ∩ B) = 1/2 * 2/3 = 1/3
Additive Law of Probability
Probability of union of events P(A U B) = P(A) + P(B) - P(A ∩ B)
If A and B are mutually exclusive P(A U B) = P(A) + P(B)
Law of Total Probability
Partitions: If S is the Sample space of the experiment and given k disjoint events B1, B2….BK
we can say that these events form the Partition of S
Law of Total Probability:
Suppose B1….Bk are the partitions over the space S and P(Bj) > 0 for j= 1…k
Then for every A in S
We can have a conditional version
Example:
A player plays a game where the final score is somewhere from 1 to 50 (equally likely)
Players score after first game is X
Player continues to play until they obtain another score Y such that Y>X
What is P(Y=50)
Lets define B i = P(X=i), the value is likely to be any number i, i+1 …. , 50 and Each of these (51-i) values for Y are equally likely
Bayes Theorem
This describes probability of an event bas on prior knowledge of conditions that might be releate to the event
P(B|A) P(A)
P(A|B) = ------------
P(B)
There is an extended form of Bayes theorem for partitions
Let events B1,….., B k form a partition set over S such that P(B j) > 0 for j = 1…k abd let A be the event such that P(A)>0 Then for i = 1..K
Conditional version
Example:
You can test a person for the presence or absence of particular disease
The test is 90% reliable
If the Person has the disease P(test positive) = 0.9
If the person does not have the disease P(test positive) = 0.1
Chance of having this disease is 1 in 10000
You have tested positive. What is the probability that you actually have a disease
Solution:
Let B1 denote you have the disease and B2 denote that you do not
B1 and B2 will for partition
Let A denote the event that the response to test is positive
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