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
Suppose that $A, B, C$ are events such that $P\,(A) = P\,(B) = P\,(C) = \frac{1}{4},\,P\,(AB) = P\,(CB) = 0,\,P\,(AC) = \frac{1}{8},$ then $P\,(A + B) = $
An event has odds in favour $4 : 5$, then the probability that event occurs, is
$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A ^{\prime} \cap B ^{\prime}\right)$.
The probability that $A$ speaks truth is $\frac{4}{5}$, while this probability for $B$ is $\frac{3}{4}$. The probability that they contradict each other when asked to speak on a fact
Fill in the blanks in following table :
$P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
$0.35$ | ........... | $0.25$ | $0.6$ |