## 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
``````
• Are A and C independent?
``````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
``````

• 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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