➢ 𝑃(Union of mutually exclusive events)= ∑ (Probability of events)
Theorem If A and B are two events associated with a random experiment, then prove that
(i) 𝑃(𝐴 ∩ 𝐵̅) = P(only A) = P(A) − P(A ∩ B)
(ii) P(𝐵 ∩ A̅) = P(only B) = P(B) − P(A ∩ B)
Proof (i) By Distributive property of sets, . (A ∩ B) ∪ (A ∩ B̅) = A ∩ (B ∪ B̅) = A ∩ S = A . (A ∩ B) ∩ (A ∩ B̅) = A ∩ (B ∩ B̅) = A ∩ 𝜙 = 𝜙 Therefore, the events A ∩ B and A ∩ B̅ are mutually exclusive whose union is A.
Therefore, P(A) = P[(A∩B)∪(A∩B̅)] P(A) = P(A∩B)+P(A∩B̅)
Therefore, P(A∩B̅)=P(A)-P(A∩B) That is, P(A∩B̅)=P(only A)=P(A)-P(A∩B)
(ii) By Distributive property of sets, . (A∩B)∪(A̅∩B)= (A∪A̅)∩B=S∩B=B . (A∩B)∩(A̅∩B)= (A∩A̅)∩B=𝜙∩B=𝜙 Therefore, the events A ∩ B and A̅ ∩ B are mutually exclusive whose union is B.
P(B)=P[(A∩B)∨(A̅∧B)] P(B)=P(A∧B)+P(A̅∧B)
Therefore, P(A̅∧B)=P(B)-P(A∧B) That is, P(A̅∧B)=P(only B)=P(B)-P(A∧B)
Progress Check
. A ∧ B and A̅ ∧ B are events. . P(𝐴̅ ∧ B)= .
. If A and B are mutually exclusive events then P(A ∧ B)= . . If P(A)= .