$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A ^{\prime} \cap B ^{\prime}\right)$.
It is given that $P ( A )=0.54$, $P ( B )=0.69$, $P (A \cap B)=0.35$
$A^{\prime} \cap B^{\prime}=(A \cup B)^{\prime}$ [by De Morgan's law]
$\therefore P \left(A^{\prime} \cap B^{\prime}\right)$ $= P (A \cup B)^{\prime}=1- P (A \cup B)=1-0.88=0.12$
Let $A$ and $B$ be events for which $P(A) = x$, $P(B) = y,$$P(A \cap B) = z,$ then $P(\bar A \cap B)$ equals
In a class of $60$ students, $30$ opted for $NCC$ , $32$ opted for $NSS$ and $24$ opted for both $NCC$ and $NSS$. If one of these students is selected at random, find the probability that The student has opted neither $NCC$ nor $NSS$.
If $A$ and $B$ are two mutually exclusive events, then $P\,(A + B) = $
In a hostel, $60 \%$ of the students read Hindi newspaper, $40 \%$ read English newspaper and $20 \%$ read both Hindi and English newspapers. A student is selected at random. If she reads Hindi newspaper, find the probability that she reads English newspaper.
Given two independent events $A$ and $B$ such $P(A)=0.3,\,P(B)=0.6 .$ Find $P($ neither $A$or $B)$