Let $X$ and $Y$ are two events such that $P(X \cup Y=P)\,(X \cap Y).$
Statement $1:$ $P(X \cap Y' = P)\,(X' \cap Y = 0).$
Statement $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$
Statement $1$ is false, Statement $2$ is true.
Statement $1$ is true, Statement $2$ is true,Statement $2$ is not a correct explanation of Statement $1$ .
Statement $1$ is true, Statement $2$ is false
Statement $1$ is true, Statement $2$ is true;Statement $2$ is a correct explanation of Statement $1.$
If from each of the three boxes containing $3$ white and $1$ black, $2$ white and $2$ black, $1$ white and $3$ black balls, one ball is drawn at random, then the probability that $2$ white and $1$ black ball will be drawn is
If $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3,$ then $P(\bar A) + P(\bar B) = $
The probability that at least one of $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$, then $P(A') + P(B') = $
Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined $P ( A )=0.5$, $ P ( B )=0.4$, $P ( A \cap B )=0.8$
Given that the events $A$ and $B$ are such that $P(A)=\frac{1}{2}, P(A \cup B)=\frac{3}{5}$ and $\mathrm{P}(\mathrm{B})=p .$ Find $p$ if they are mutually exclusive.