Mathematics CONDITIONAL PROBABILITY

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CONDITIONAL PROBABILITY

Conditional Probability :

Let `A` and `B` be any two events associated with a random experiment. The probability of occurrence of event `A` when the event `B` has already occurred is called the conditional probability of `A` when `B` is given and is denoted as `P(A//B)`. The conditional probability `P(A//B)` is meaningful only when `P(B) ne 0`, i.e., when `B` is not an impossible event.

By definitiion ,

`P(A/B)=`Probability of occurrence of event `A` when the event `B` as already occurred

`=(text(Number of cases favourable to B which are also favourable to A))/(text(Number of cases favourable to B))`

`:. P(A/B)=(text(Number of cases favourable to A cap B))/(text(Number of cases favourable to B))`

Also, `P(A/B)=((text(Number of cases favourable to A cap B))/(text(Number of cases in the sample space)))/((text(Number of cases favourable to B))/(text(Number of cases in the sample space)))`

`:. P(A/B)=(P(A cap B))/(P(B))`, provided `P(B) ne 0`.

Similarly, we have

`P(B/A)=(P(A cap B))/(P(A))`, provided `P(A) ne 0`.

Let `A` and `B` be any two events associated with a random experiment. The probability of occurrence of event `A` when the event `B` has already occurred is called the conditional probability of `A` when `B` is given and is denoted as `P(A//B)`. The conditional probability `P(A//B)` is meaningful only when `P(B) ne 0`, i.e., when `B` is not an impossible event.

By definitiion ,

`P(A/B)=`Probability of occurrence of event `A` when the event `B` as already occurred

`=(text(Number of cases favourable to B which are also favourable to A))/(text(Number of cases favourable to B))`

`:. P(A/B)=(text(Number of cases favourable to A cap B))/(text(Number of cases favourable to B))`

Also, `P(A/B)=((text(Number of cases favourable to A cap B))/(text(Number of cases in the sample space)))/((text(Number of cases favourable to B))/(text(Number of cases in the sample space)))`

`:. P(A/B)=(P(A cap B))/(P(B))`, provided `P(B) ne 0`.

Similarly, we have

`P(B/A)=(P(A cap B))/(P(A))`, provided `P(A) ne 0`.

Properties of conditional probability

Let E and F be events of a sample space S of an experiment, then we have

`text(Property : 1)`

`P(S|F) = P(F|F) = 1`

We know that

`P(S|F) = (P(S nn F)) /(P(F)) = (P(F) )/ (P(F)) =1 `

Also `P(F|F) = (P(E nn F)) / (P(F)) = (P(F)) /(P(F)) = 1`

`P(S|F) = P(F|F) = 1`

`text(Property : 2)`

If A and B are any two events of a sample space S and F is an event of S such that `P(F) ≠ 0,` then

`P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F)`

In particular, if A and B are disjoint events, then `P((A∪B)|F) = P(A|F) + P(B|F)`

`P((A∪B)|F) = P(A|F) + P(B|F)`

we have

`P((A∪B)|F) = (P[(A uu B) nn F])/ (P(F))`

`= (P[(A nn F) uu (B nn F)])/(P(F))` (by distributive law of union of sets over intersection)

` = (P(A nn F)+P(B nn F) – P(A nn B nn F))/(P(F))`

` = (P(A nn F))/(P(F)) + (P(BnnF))/(P(F)) + (P(A nn B))/( P(F))`

`= P(A|F) + P(B|F) – P((A ∩B)|F)`

When A and B are disjoint events, then `P((A ∩ B)|F) = 0`

`⇒ P((A ∪ B)|F) = P(A|F) + P(B|F)`

`text(Property: 3)`

` P(E′|F) = 1 − P(E|F)`

From Property 1, we know that `P(S|F) = 1`

=> `P(E ∪ E′|F) = 1` `\ \ \ \ because S = E ∪ E′`

=> `P(E|F) + P (E′|F) = 1` since E and E′ are disjoint events

`P(E′|F) = 1 − P(E|F)`

Let E and F be events of a sample space S of an experiment, then we have

`text(Property : 1)`

`P(S|F) = P(F|F) = 1`

We know that

`P(S|F) = (P(S nn F)) /(P(F)) = (P(F) )/ (P(F)) =1 `

Also `P(F|F) = (P(E nn F)) / (P(F)) = (P(F)) /(P(F)) = 1`

`P(S|F) = P(F|F) = 1`

`text(Property : 2)`

If A and B are any two events of a sample space S and F is an event of S such that `P(F) ≠ 0,` then

`P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F)`

In particular, if A and B are disjoint events, then `P((A∪B)|F) = P(A|F) + P(B|F)`

`P((A∪B)|F) = P(A|F) + P(B|F)`

we have

`P((A∪B)|F) = (P[(A uu B) nn F])/ (P(F))`

`= (P[(A nn F) uu (B nn F)])/(P(F))` (by distributive law of union of sets over intersection)

` = (P(A nn F)+P(B nn F) – P(A nn B nn F))/(P(F))`

` = (P(A nn F))/(P(F)) + (P(BnnF))/(P(F)) + (P(A nn B))/( P(F))`

`= P(A|F) + P(B|F) – P((A ∩B)|F)`

When A and B are disjoint events, then `P((A ∩ B)|F) = 0`

`⇒ P((A ∪ B)|F) = P(A|F) + P(B|F)`

`text(Property: 3)`

` P(E′|F) = 1 − P(E|F)`

From Property 1, we know that `P(S|F) = 1`

=> `P(E ∪ E′|F) = 1` `\ \ \ \ because S = E ∪ E′`

=> `P(E|F) + P (E′|F) = 1` since E and E′ are disjoint events

`P(E′|F) = 1 − P(E|F)`

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