If ${A_1},\,{A_2},...{A_n}$ are any $n$ events, then
$P\,({A_1} \cup {A_2} \cup ... \cup {A_n}) = P\,({A_1}) + P({A_2}) + ... + P\,({A_n})$
$P\,({A_1} \cup {A_2} \cup ... \cup {A_n}) > P\,({A_1}) + P({A_2}) + ... + P\,({A_n})$
$P\,({A_1} \cup {A_2} \cup ... \cup {A_n}) \le P\,({A_1}) + P({A_2}) + ... + P\,({A_n})$
None of these
$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A \cap B ^{\prime}\right)$ .
Three coins are tossed simultaneously. Consider the event $E$ ' three heads or three tails', $\mathrm{F}$ 'at least two heads' and $\mathrm{G}$ ' at most two heads '. Of the pairs $(E,F)$, $(E,G)$ and $(F,G)$, which are independent? which are dependent ?
The probability of happening at least one of the events $A$ and $B$ is $0.6$. If the events $A$ and $B$ happens simultaneously with the probability $0.2$, then $P\,(\bar A) + P\,(\bar 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 $B).$
The probability that a man will be alive in $20$ years is $\frac{3}{5}$ and the probability that his wife will be alive in $20$ years is $\frac{2}{3}$. Then the probability that at least one will be alive in $20$ years, is