If $\mathrm{A}$ and $\mathrm{B}$ are two events such that $P(\displaystyle \mathrm{A}\cup B)\geq\frac{3}{4}$ and $\displaystyle \frac{1}{8}\leq P(\mathrm{A}\cap B)\leq\frac{3}{8}$ then

Probability of any event $x$ lies

If $A and B$ are two events such that $P\left( A\cup B \right) +P\left( A\cap B \right) =\dfrac { 7 }{ 8 }$ and $P\left(A\right) = 2P\left(B\right), then$P\left(A\right) = \(

The probability that $A, B$ and $C$ can solve a problem are $\displaystyle \frac{4}{5}, \frac{2}{3}$ and $\dfrac{3}{7}$ respectively. The probability of the problem being solved by $A$ and $B$ is $\dfrac{8}{15}$, $B$ and $C$ is $\dfrac{2}{7}$, $A$ and $C$ is $\dfrac{12}{35}$. The probability of the problem being solved by all the three is $\dfrac{8}{35}$.

Find the probability that the problem can be solved by at least one of them.

If $A$ and $B$ are two independent events such that $P(A)=\dfrac{1}{3}$ and $P(B)=\dfrac{3}{4}$, then $p\left\{\dfrac{B}{(A\cup B)}\right\}$