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Let $S$ be a sample space and two mutually exclusive events $A$ and $B$ be such that $A \cup B = S$. If $P(.)$ denotes the probability of the event, the maximum value of $P(A)P(B)$ is_____.

1/2 * 1/2 =1/4

P(A) + P(B) = 1, since both are mutually exclusive and A ∪ B = S.
When sum is a constant, product of two numbers becomes maximum when they are equal. So, P(A) = P(B) = 1/2
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Suppose E:Even, O:Odd ,S:Sample Space of Natural Numbers

Example of such an mutually exclusive event is P(E)=P(O)=1/2

P(E)+P(O)=S(All Natural Numbers)=1

So Ans is

P(E)*P(O)= $\frac{1}{2}$* $\frac{1}{2}$=$\frac{1}{4}$

Sample Space(S) - A set of all possible outcomes/events of a random experiment. Mutually Exclusive Events - Those events which can't occur simultaneously.   P(A)+P(B)+P(A∩B)=1   Since the events are mutually exclusive, P(A∩B)=0. Therefore, P(A)+P(B)=1   Now, we now that AM >= GM So, (P(A)+P(B))/2 >= sqrt(P(A)*P(B))   P(A)*P(B) <= 1/4
Hence max(P(A)*P(B)) = 1/4.
We can think of this problem as flipping a coin, it has two mutually exclusive events ( head and tail , as both can't occur simultaneously). And sample space S = { head, tail }   Now, lets say event A and B are getting a "head" and "tail" respectively. Hence, S = A U B.   Therefore, P(A) = 1/2 and P(B) = 1/2.   And, P(A).P(B) = 1 /4 = 0.25.   Hence option B is the correct choice.