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 $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$
Let $A$ and $B$ are two events and $P(A') = 0.3$, $P(B) = 0.4,\,P(A \cap B') = 0.5$, then $P(A \cup B')$ is
A die is tossed thrice. Find the probability of getting an odd number at least once.
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') = $
$A$ and $B$ are events such that $P(A)=0.42$, $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P ($ not $B).$