## 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

- General Examples
- 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

- General Events:

- Dependent Events:
- 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
```

## 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