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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_____.
in Probability by Veteran (104k points) | 2.2k views
Use concept -->If perimeter is fix then square has maximum area. :)

4 Answers

+24 votes
Best answer
$\dfrac{1}{2}\times \dfrac{1}{2} =\dfrac{1}{4}$

$P(A) + P(B) = 1,$ since both are mutually exclusive and $A\cup B = S.$
When sum is a constant, product of two numbers becomes maximum when they are equal.

So,$ P(A) = P(B) =\dfrac{1}{2}.$
by Junior (883 points)
edited by
Given that A and B are mutually exclusive events than

P(A)+P(B)=1. now let say probability of A = x than probability of B = 1-x.

When need maximum probability P(A)*P(B) CAN HAVE


differentiate wrt to x:-





Since A U B =S is given, therefore A and B  are dependent events. So P(A) P(B) != 0. Am i right?
it is given that A and B are Mutually Exclusive, so $P(A \cap B)=\phi$

and also given $A \cup B=S$ means A and B are collectively exhaustive.
+10 votes

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)= $\frac{1}{2}$* $\frac{1}{2}$=$\frac{1}{4}$

by Boss (23.7k points)
+6 votes
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.
by Loyal (9.6k points)
edited by
Nice solution by using the property AM >= GM.
+4 votes
When P(A)P(B)  is maximum A,B together should complete whole universal space( P(A∪B)=1,  also in the question it is mentioned that A∪B=S ,so P(A U B)=1) and P(A),P(B) not equal to 0

It is given that A,B are mutually exclusive events(P(A∪B)= P(A) + P(B) ). So, if we let P(A)=x then,

P(A∪B)= P(A) + P(B) or, 1= x +P(B) or, P(B)=1-x.  Hence, P(A)P(B)= x.(1-x)

derivative of x(1-x)= 1-2x,  So,  when P(A)P(B) is maximum, 1-2x=0 or, x=1/2, Hence x(1-x)= 1/2 * 1/2 = 1/4=0.25
by Active (2.8k points)
edited by

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