Question

Is conditional probability independent or mutually exclusive? Or neither? Please explain with examples! It would mean the world!

Answer 1

If occurence of one events excludes the occurence of other event then they are mutually exclusive events.

When two events are mutually exclusive then the intersection of the events is 0.

when two events are independent then their intersection is the product of the individual probability of events.

Conditional probability is the probability of an event when another event occurs.

It is defined as the probability of occurence of event A when event B occured.

A conditional probability may be independent but not mutually exclusive because in mutually exclusive occurence of event result in not occurence of other event.

ie., if A and B are mutually exclusive then if event A occurs then event B does not occurs and viceversa.

if this there is no scope for condition of occurence of a event affects the other.

EXAMPLES:

Event 1: Probability that the flight is on time today

Event 2: Probability that the flight is delayed today

these two events are mutually exclusive events because if the flight is on time today then there is no chance of delay in the arrival. In the same manner if the flight is delayed today then there is no scope of arriving on time . Here the conditional probability does not occur as there is no condition of happening of both the events.

EXAMPLE 2

when a die is rolled twice

Event 1 : Probabilty of getting an even number in the first toss

Event 2 : Probability of getting 2 in the second toss

When a die is rolled the probability of occurence of any number is independent of the other.The first toss is independent of the second toss. though the events are independent there can be a conditional probability which is equal to the probability of that event itself.

P(Event1|Event2) =P(event1)

P(event2|Event1)= P(Event2)

Probabilty of getting an even number when a six sided die is rolled =3/6 =1/2

Probability of getting 2 =1/6

probability of getting 2 in second tossgiven that the rolled number is even in the first toss = P(Event2|Event1) =P(event2) =1/6