# show that prob (AUB)=prob(A)+prob(B)-prob(AnB).in particular ,if A and B are independent events,then prob(AuB)=prob(A)+prob(B)-prob(A)*prob(B) Consider a probability space S with probability...

show that prob (AUB)=prob(A)+prob(B)-prob(AnB).in particular ,if A and B are independent events,then prob(AuB)=prob(A)+prob(B)-prob(A)*prob(B)

Consider a probability space S with probability function prob. Let A, B is element of s be two events.

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If A and B are two events of the sample space;Let B' be complementry event of B, By drawin a venn diagran you can clearly see that,

(AnB')n(B) = null

Therefore (AnB') and B are mutually exclusive events.

So, P[(AnB')U(B)]= P(AnB')+P(B);

But, from venn diagram you can observe that;

(AnB')U(B)=(AUB)

Therefore;

P[AUB]= P(AnB')+P(B)............(1)

Let us consider AnB';

If A and B are two events of the sample space;Let B' be complementry event of B, By drawin a venn diagran you can clearly see that,

(AnB')n(AnB)={}=Null

Therefore, (AnB') and (AnB) are mutually exclusive events.

P[(AnB')U(AnB)]=P(AnB')+P(AnB)

But, from the venn diagram (AnB')U(AnB)=A

Therefore;

P(A)=P(AnB')+P(AnB)

P(AnB')=P(A)-P(AnB).........(2)

By substituting From (2) to (1);

**P[AUB]= P(A)+P(B)-P(AnB)...........(i)**

**By the definition of conditional probability,**

**P(A\B)=P(AnB)/P(B)**

**Therefore P(AnB)=P(A\B)*P(B)........(3)**

**If A and B are independent events occurance of B doesn't affect the occurance of A.**

**Therefore;**

**P(A\B)**=P(A)

**Subsitituting to (3);**

** P(AnB)=P(A\B)*P(B)**=**P(A)*P(B)**

**By subsituting to (i);**

**P[AUB]= P(A)+P(B)-P(A)*(B)**

**Think you have knowledge on venn diagrams.**

If event A occurs B will not occur then P(A)=1 , P(B)=0 then P(A)*P(B)=0 vice versa.

If A and B are independant events then there will be no common elements. Hence A & B will not occur simultaneously so P(A)*P(B) will be zero . We know that P(AuB)=P(A)+P(B) if A and B are independant . Since P(A)*P(B)=0 we can write P(AuB)=P(A)+P(B)-P(A)*P(B) .