If $A$ and $B$ are any two events, then $P(A \cup B) = $
$P(A) + P(B)$
$P(A) + P(B) + P(A \cap B)$
$P(A) + P(B) - P(A \cap B)$
$P(A)\,\,.\,\,P(B)$
If the odds in favour of an event be $3 : 5$, then the probability of non-occurrence of the event is
$A$ and $B$ are events such that $P(A)=0.42$, $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P (A$ or $B).$
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 $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$
If $E$ and $F$ are events such that $P ( E )=\frac{1}{4}$, $P ( F )=\frac{1}{2}$ and $P(E$ and $F )=\frac{1}{8},$ find : $P ( E$ or $F )$
If $A$ and $B$ are any two events, then the probability that exactly one of them occur is