A problem in mathematics is given to three students whose chances of solving it are $\dfrac{1}{3}$, $\dfrac{2}{7}$ and $\dfrac{3}{8}$ respectively. What is the probability of the problem will be solved?

Hint: First, we will assign the name and the corresponding events of the 3 students solving the questions. Then, we will find the complementary events of those probabilities. In the final step, by using the condition of intersection of complements, we will find the probability that the problem will be solved.

Complete step by step answer:
Here, there is a mathematics question which needs to be solved by 3 students, let us say Student L, Student M and Student N and the probabilities of the students solving the question is $\dfrac{1}{3}$, $\dfrac{2}{7}$ and $\dfrac{3}{8}$ respectively.
First, let us assign the events to the students solving the mathematics question.
Let X be the event that student L solves the question, therefore the probability of L solving the question, $P\left( X \right)=\dfrac{1}{3}$ and their complementary is $P\left( \overline{X} \right)=\dfrac{2}{3}$.
Let Y be the event that student M solves the question, therefore the probability of M solving the question, $P\left( Y \right)=\dfrac{2}{7}$ and their complementary is $P\left( \overline{Y} \right)=\dfrac{5}{7}$.
Let Z be the event that student N solves the question, therefore the probability of N solving the question, $P\left( Z \right)=\dfrac{3}{8}$ and their complementary is $P\left( \overline{Z} \right)=\dfrac{5}{8}$.
To find probability that the problem will be solved we need to solve it by intersection of complements,
We know, by intersection of complements
$P\left( X\cup Y\cup Z \right)$$=1-P\left( \overline{X}\cap \overline{Y}\cap \overline{Z} \right)$
Now, here the 1 represents the total probability of success and failure combined, and \[P\left( \overline{X}\cap \overline{Y}\cap \overline{Z} \right)\] represents that out of all those 3 students none of them have solved that question, so when we solve the above expression, we will get the probability of the problem being solved, $P\left( X\cup Y\cup Z \right)$. We know, that events $X$, $Y$ and $Z$ are independent events, therefore, $\overline{X}$, $\overline{Y}$and $\overline{Z}$ are also independent events respectively.
Since, we know the intersection of the independent sets is equal to multiplication of their respective probabilities, we get
$P\left( X\cup Y\cup Z \right)$$=1-P\left( \overline{X} \right)\cdot P\left( \overline{Y} \right)\cdot P\left( \overline{Z} \right)$
$\begin{align}
  &\Rightarrow P\left( X\cup Y\cup Z \right) =1-\left( \dfrac{2}{3} \right)\cdot \left( \dfrac{5}{7} \right)\cdot \left( \dfrac{5}{8} \right) \\
 &\Rightarrow P\left( X\cup Y\cup Z \right) =1-\left( \dfrac{1}{3} \right)\cdot \left( \dfrac{5}{7} \right)\cdot \left( \dfrac{5}{4} \right) \\
 &\Rightarrow P\left( X\cup Y\cup Z \right) =1-\left( \dfrac{1\times 5\times 5}{3\times 7\times 4} \right) \\
 &\Rightarrow P\left( X\cup Y\cup Z \right) =1-\left( \dfrac{25}{84} \right) \\
 &\Rightarrow P\left( X\cup Y\cup Z \right) =\dfrac{1\times 84}{1\times 84}-\dfrac{25}{84} \\
 &\Rightarrow P\left( X\cup Y\cup Z \right) =\dfrac{84-25}{84} \\
 &\Rightarrow P\left( X\cup Y\cup Z \right) =\dfrac{59}{84} \\
\end{align}$

Hence, the probability that the problem will be solved is $\dfrac{59}{84}$.

Note: Normally, when there will be a set of data given to us instead of the value of the probability of either success or failure, then the probability of an event will be the ratio of the number of cases in the favour of the condition to the total number of cases. Complementary to an event E is given by, $P\left( \overline{\text{E}} \right)=1-P\left( \text{E} \right)$.