$A$ 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]
$\therefore P \left(A^{\prime} \cap B^{\prime}\right)$ $= P (A \cup B)^{\prime}=1- P (A \cup B)=1-0.88=0.12$
The probability of happening at least one of the events $A$ and $B$ is $0.6$. If the events $A$ and $B$ happens simultaneously with the probability $0.2$, then $P\,(\bar A) + P\,(\bar B) = $
If $P\,(A) = 0.4,\,\,P\,(B) = x,\,\,P\,(A \cup B) = 0.7$ and the events $A$ and $B$ are mutually exclusive, then $x = $
If $A, B, C$ are three events associated with a random experiment, prove that
$P ( A \cup B \cup C ) $ $= P ( A )+ P ( B )+ P ( C )- $ $P ( A \cap B )- P ( A \cap C ) $ $- P ( B \cap C )+ $ $P ( A \cap B \cap C )$
One card is drawn from a pack of $52$ cards. The probability that it is a queen or heart is
The probabilities of three events $A , B$ and $C$ are given by $P ( A )=0.6, P ( B )=0.4$ and $P ( C )=0.5$ If $P ( A \cup B )=0.8, P ( A \cap C )=0.3, P ( A \cap B \cap$ $C)=0.2, P(B \cap C)=\beta$ and $P(A \cup B \cup C)=\alpha$ where $0.85 \leq \alpha \leq 0.95,$ then $\beta$ lies in the interval