Question

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

• A. 0.25

Answer 1

The correct option is A 0.25
$\because$ A & B are mutually exclusive so $P(A\cap B)=0$
Now $(A\cup B)=S$
$\Rightarrow P(A\cup B)=P\left(S\right)=1$
$\Rightarrow P\left(A\right)+P\left(B\right)=1$

Let P(A) = x & P(B) = y then we have x + y = 1

Now P(A)P(B) = x.y

Max [P(A).P(B)] will occur only when x = y
Hence, Max[P(A).P(B)]
$=Max\left(xy\right)=x^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$