Let $A$ and $B $ be two events such that  $P\left( {\overline {A \cup B} } \right) = \frac{1}{6}\;,P\left( {A \cap B} \right) = \frac{1}{4}$ and $P\left( {\bar A} \right) = \frac{1}{4}$ where $\bar A$ stands for the complement of the event $A$. Then the events $A$ and$B$ are

$A$ 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 of solving a question by three students are $\frac{1}{2},\,\,\frac{1}{4},\,\,\frac{1}{6}$ respectively. Probability of question is being solved will be

$A$ 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 at least one of the events $A$ and $B$ occurs is $3/5$. If $A$ and $B$ occur simultaneously with probability $1/5$, then $P(A') + P(B')$ is