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

Option: 1

0.25


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\because \mathrm{A\, \& \, B}  are mutually exclusive so \mathrm{P}(A \cap B)=0

\mathrm{\text { Now } (A \cup B) =S }
\mathrm{\\ \Rightarrow \quad P(A \cup B) =P(S)=1 }
\mathrm{\Rightarrow \quad P(A)+P(B) =1 }

\mathrm{\text { Let } \quad P(A) =x \, \&\, P(B) }
                         \mathrm{=y \text { then we have } x+y=1}

\mathrm{Now \, P(A) P(B)=x \cdot y}
\mathrm{Max [P(A) \cdot P(B)]} will occur only when \mathrm{x=y}

Hence,
\mathrm{\operatorname{Max}[\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})]}
\mathrm{=\operatorname{Max}(x y)=x^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}}

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