Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let be a sample space and two mutually exclusive events <img alt="\mathrm{A}" src="https://entranceco

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}}

Posted by

chirag

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question