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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_____.

Use concept -->If perimeter is fix then square has maximum area. :)
Use concept of arithmetic mean(AM) and geometric mean(GM)

for any two positive numbers

$AM\geq GM$

$AM = (P(A)+P(B))/2$

$GM = \sqrt{P(A)*P(B)}$

$(P(A)+P(B))/2 \geq \sqrt{P(A)*P(B)}$

Put $P(A)+P(B)=1$

So $P(A)*P(B)\leq 1/4$

A U B = S → P(A U B) = P(S) = 1

P(A U B) = P(A) + P(B) – P(A ${\cap }^{}$ B)  [ P(A ${\cap }^{}$ B) = Phi,as they are mutually exclusive events]

So, P(AUB) = P(A) + P(B) = 1

P(A)xP(B) = P(A)x(1-P(A)) [ as P(A) + P(B) = 1]

Put, P(A) = Z , P(A)xP(B) = P(A)x(1-P(A)) = Zx(1-Z)

Maximum value of P(A)xP(B) = Maximum value of Zx(1-Z) = f(Z)

The maximum value of f(Z) will occur when f(Z) =0  [f(Z) =  Derivation of f(Z) with respect to Z]

Now f(Z) = 1-2xZ = 0 → Z = 1/2

So, P(A)xP(B) = Zx(1-Z) will be maximum when Z = ½

So, Maximum value of P(A)xP(B) = Zx(1-Z) = (1/2)x(1-1/2) = ¼ = 0.25

$\dfrac{1}{2}\times \dfrac{1}{2} =\dfrac{1}{4}$

$P(A) + P(B) = 1,$ 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

Visual Representation for Clarity:- Since $A \cup B = S$. So, we can partition $S$ into two parts $A$ and $B$.

We know that max value of product of $P(A)$ and $P(B)$ can happen only if they are equal and since $P(A)+P(B) = S$, so $P(A) = P(B) = 0.5$.

[If they were like $70\%$ and $30\%$, then $P(A) * P(B) = 0.7*0.3 = 0.21$ which is $< (0.5)*(0.5) = (0.25)$]

( Just for clarity purpose )

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Here, P(A) + P(B) = 1

Let S = P(A).P(B)

Therefore, S = P(A)(1 – P(A))

S = P(A) – P(A)^2

Differentiating, both sides w.r.t. P(A) –

dS/dx = 1 – 2P(A)

for maximize the value of s,  dS/dx = 0

1 – 2P(A) = 0

P(A) = ½

therefore, S = ½ * ½ = ¼

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.

Nice solution by using the property AM >= GM.
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