Find $P\left( A\cup B \right)$ when $2P\left( A \right)=P\left( B \right)=\dfrac{5}{13}$ and $P\left( \dfrac{A}{B} \right)=\dfrac{2}{5}$.

Hint: We will apply the formula $P\left( \dfrac{A}{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$ and $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$ in order to solve the question. As we are not given the direct value of P(A) so we will use multiplication here to solve it further.

Complete step-by-step answer:
As we are given that $2P\left( A \right)=P\left( B \right)=\dfrac{5}{13}$ so, we will find out the value of P(A) from this equation. This can be done by considering $2P\left( A \right)=\dfrac{5}{13}$.
After multiplying this equation by 2 on both the sides then, we will get $\dfrac{2P\left( A \right)}{2}=\dfrac{5}{13\left( 2 \right)}$.
Using simplification by division we will get $P\left( A \right)=\dfrac{5}{26}$.
Now we have the values of $P\left( A \right)=\dfrac{5}{26}$ and $P\left( B \right)=\dfrac{5}{13}$. Also, we have that $P\left( \dfrac{A}{B} \right)=\dfrac{2}{5}$ so, we are clear here to use the formula $P\left( \dfrac{A}{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$. We already have the value of P(B) but we need to find the value of the intersection of the probability of A and B. This can be done by substitution of the values of $P\left( \dfrac{A}{B} \right)=\dfrac{2}{5}$ and $P\left( B \right)=\dfrac{5}{13}$. Thus, we now have
$\begin{align}
  & P\left( \dfrac{A}{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)} \\
 & \Rightarrow \dfrac{2}{5}=\dfrac{P\left( A\cap B \right)}{\dfrac{5}{13}} \\
\end{align}$
At this step we will use the cross multiplication here. We will do so by taking the fraction $\dfrac{5}{13}$ to the left side of the equation. Therefore, now we have $\dfrac{2}{5}\times \dfrac{5}{13}=P\left( A\cap B \right)$. Now we will cancel 5 by 5 as the product of 5 and 1 is 5. Thus, we will get $\dfrac{2}{1}\times \dfrac{1}{13}=P\left( A\cap B \right)$. With the multiplication we now have $P\left( A\cap B \right)=\dfrac{2}{13}$.
Now we will use the formula given by $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$. As we have the values of $P\left( A \right)=\dfrac{5}{26}$ and $P\left( B \right)=\dfrac{5}{13}$ also, the value of $P\left( A\cap B \right)=\dfrac{2}{13}$ then after substituting these values in the formula $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$ will result into $P\left( A\cup B \right)=\dfrac{5}{26}+\dfrac{5}{13}-\dfrac{2}{13}$. Now we will take the lcm of 26 and 13. We will get the lcm as 26 ony. Therefore, we will get
$\begin{align}
  & P\left( A\cup B \right)=\dfrac{5+10-4}{26} \\
 & \Rightarrow P\left( A\cup B \right)=\dfrac{15-4}{26} \\
 & \Rightarrow P\left( A\cup B \right)=\dfrac{11}{26} \\
\end{align}$
Hence, the value of $P\left( A\cup B \right)$ is $\dfrac{11}{26}$.

Note: As we have so many values in this equation one should be focused while substituting the values in the formulas. If we were asked to find the value of $P\left( A\cup B \right)$ in terms of decimal then we can do this by dividing 11 by 26. The value after this will get converted into 0.42, which is up to 2 decimal places. If in case we need to verify that whether the value of the probability is correct then we can do this by having the value in decimal. If the value is between 0 and 1 then the chances of the answer to be correct will be increased.