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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_____.
asked in Probability by Veteran (106k points) | 1.1k views
Use concept -->If perimeter is fix then square has maximum area. :)

4 Answers

+12 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}.$
answered by Junior (783 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

P(A)*P(B)=x*(1-x).

differentiate wrt to x:-

$\frac{d}{dx}$(x*(1-x)=x(-1)+(1-x)=0

1-x=x

1=2x

x=1/2=0.5

P(A)*P(B)=0.5*0.5=0.25
+3 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)= $\frac{1}{2}$* $\frac{1}{2}$=$\frac{1}{4}$

answered by Veteran (22.8k points)
+2 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.
answered by Boss (7.8k points)
Nice solution by using the property AM >= GM.
+1 vote
When P(A)P(B)  is maximum A,B together should complete whole universal space( P(A∪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
answered by Active (1.2k points)


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