Two dice are thrown independently. Let $A$ be the event that the number appeared on the $1^{\text {st }}$ die is less than the number appeared on the $2^{\text {nd }}$ die, $B$ be the event that the number appeared on the $1^{\text {st }}$ die is even and that on the second die is odd, and $C$ be the event that the number appeared on the $1^{\text {st }}$ die is odd and that on the $2^{\text {nd }}$ is even. Then
the number of favourable cases of the event $(A \cup B) \cap C$ is $6$
$A$ and $B$ are mutually exchusive
The number of favourable cases of the events $A , B$ and $C$ are $15,6$ and $6$ respectively
$B$ and $C$ are independent
If $P\,({A_1} \cup {A_2}) = 1 - P(A_1^c)\,P(A_2^c)$ where $c$ stands for complement, then the events ${A_1}$ and ${A_2}$ are
Let $A$,$B$ and $C$ be three events such that $P\left( {A \cap \bar B \cap \bar C} \right) = 0.6$, $P\left( A \right) = 0.8$ and $P\left( {\bar A \cap B \cap C} \right) = 0.1$, then the value of $P$(atleast two among $A$,$B$ and $C$ ) equals
Probability of solving specific problem independently by $A$ and $B$ are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that exactly one of them problem
Let $A$ and $B $ be two events such that $P\left( {\overline {A \cup B} } \right) = \frac{1}{6}\;,P\left( {A \cap B} \right) = \frac{1}{4}$ and $P\left( {\bar A} \right) = \frac{1}{4}$ where $\bar A$ stands for the complement of the event $A$. Then the events $A$ and$B$ are
If $A$ and $B$ are any two events, then $P(\bar A \cap B) = $