Answer :

(i)



WE HAVE TO PROVE,


P(A)= P(AՈB) + P(AՈ)


We know A = AՈS


We can write s = A Ս A and s = B Ս B


So,


A = A Ո (B Ս B)


= (A Ո B) Ս (A Ո B)


Two events are mutually exclusive or disjoint if they cannot both occur at the same time.


(A Ո B) means A and B both occurring at the same time while (A Ո B) means A and B both occurring at the same time.


So, it is not possible that (A Ո B) and (A Ո B) occur at the same time.


Hence (A Ո B) and (A Ո B) are mutually exclusive.


When events are mutually exclusive then P (A Ո B) = 0


P[(A Ո B) Ո (A Ո B)] = 0 ….. (1)


So, A = A Ո (B Ս B)


As we know P (A Ս B) = P (A) + P(B) - P (A Ո B)


P(A) = P [(A Ո B) Ս (A Ո B)]


= P (A Ո B) + P (A Ո B) – P [(A Ո B) Ո (A Ո B)]


From (1),


P(A) = P(AՈB) + P(AՈ)


Hence proved


(ii)WE HAVE TO PROVE, P(AՍB) = P(AՈB) + P(AՈ) + P(ՈB)



AՍB means the all the possible outcomes of both A and B.


From the Venn diagram we can see,


AՍB = (AՈB) Ս (AՈ) Ս (ՈB)


Two events are mutually exclusive or disjoint if they cannot both occur at the same time.


(A Ո B) means A and B both occurring at the same time while (A Ո B) means A and B both occurring at the same time.


(ՈB) means A and B both occurring at the same time.


So, it is not possible that (A Ո B), (A Ո B) and (ՈB) occur at the same time.


Hence (A Ո B), (A Ո B) and (ՈB) are mutually exclusive.


When events are mutually exclusive then P (A Ո B) = 0


P [(A Ո B) Ո (A Ո B)] =0 ….. (1)


P [(A Ո B) Ո P(ՈB)] = 0 ….. (2)


P [(A Ո B) Ո P(ՈB)] = 0 ….. (3)


P [(A Ո B) Ո (A Ո B) Ո P(ՈB)] = 0 …. (4)


P(AՍB) = P[(AՈB) Ս (AՈ) Ս (ՈB)]


We know,


P(A B C) = P(A) + P(B) + P(C) − P(A B) P(A C) P(B C) + P(A B C)


So,


P(AՍB) = P[(AՈB) Ս (AՈ) Ս (ՈB)] = P(AՈB) + P() + P(ՈB) – P[(AՈB) (] – P[(AՈB) (ՈB)]− P[(AՈ) (ՈB)] + P [(A Ո B) Ո (A Ո B) Ո P(ՈB)]


From (1), (2), (3) and(4) we get,


P(AՍB) = P(AՈB) + P(AՈ) + P(ՈB)


Hence proved.


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