The probability of solving a question by three students are $\dfrac{1}{2}$ , $\dfrac{1}{4}$ , $\dfrac{1}{6}$ respectively. The probability of question is being solved will be
1) $\dfrac{33}{48}$
2) $\dfrac{35}{48}$
3) $\dfrac{31}{48}$
4) $\dfrac{37}{48}$

Hint: In this problem we need to calculate the probability of solving a question. We have given the probabilities of solving a question by three students. Assume the probabilities as $P\left( A \right)$ , $P\left( B \right)$ and $P\left( C \right)$ respectively. We need to calculate the probability of solving a problem which will be the value of $P\left( A\cup B\cup C \right)$ . Here the events $A$ , $B$ and $C$ are independent events, so the value of $P\left( A\cup B\cup C \right)$ will be $P\left( A\cup B\cup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-\left[ P\left( A \right).P\left( B \right)+P\left( B \right).P\left( C \right)+P\left( C \right).P\left( A \right) \right]+P\left( A \right).P\left( B \right).P\left( C \right)$. By using the above formula we will calculate the required result.

Complete step-by-step solution:
The probability of solving a problem by three students are $\dfrac{1}{2}$ , $\dfrac{1}{4}$ , $\dfrac{1}{6}$ respectively.
Let us assume that $P\left( A \right)=\dfrac{1}{2}$ , $P\left( B \right)=\dfrac{1}{4}$ and $P\left( C \right)=\dfrac{1}{6}$ .
Now the probability of solving the problem is given by $P\left( A\cup B\cup C \right)$.
Here the events $A$ , $B$ and $C$ are mutually exclusive or independent events. So the value of $P\left( A\cup B\cup C \right)$ is given by
$P\left( A\cup B\cup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-\left[ P\left( A \right).P\left( B \right)+P\left( B \right).P\left( C \right)+P\left( C \right).P\left( A \right) \right]+P\left( A \right).P\left( B \right).P\left( C \right)$
Substituting all the values we have in the above equation, then we will have
$P\left( A\cup B\cup C \right)=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}-\left[ \dfrac{1}{2}\times \dfrac{1}{4}+\dfrac{1}{4}\times \dfrac{1}{6}+\dfrac{1}{6}\times \dfrac{1}{2} \right]+\dfrac{1}{2}\times \dfrac{1}{4}\times \dfrac{1}{6}$
Simplify the above equation by using basic mathematical operations, then we will get
$\begin{align}
  & P\left( A\cup B\cup C \right)=\dfrac{6+3+2}{12}-\left[ \dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{12} \right]+\dfrac{1}{48} \\
 & \Rightarrow P\left( A\cup B\cup C \right)=\dfrac{11}{12}-\left[ \dfrac{3+1+2}{24} \right]+\dfrac{1}{48} \\
 & \Rightarrow P\left( A\cup B\cup C \right)=\dfrac{11}{12}-\dfrac{6}{24}+\dfrac{1}{48} \\
 & \Rightarrow P\left( A\cup B\cup C \right)=\dfrac{4\left( 11 \right)-6\left( 2 \right)+1}{48} \\
\end{align}$
On further simplification we will have
$\begin{align}
  & P\left( A\cup B\cup C \right)=\dfrac{44-12+1}{48} \\
 & \Rightarrow P\left( A\cup B\cup C \right)=\dfrac{33}{48} \\
\end{align}$
Hence the probability of solving a problem is $\dfrac{33}{48}$ . So option 1 is the correct answer.

Note: We can also solve the problem by using another method. In this method we will use the concept of complementary events and the sum probabilities of all events is one. We know that the sum of probabilities of solving a problem and not solving a problem is equal to one i.e.,
$\begin{align}
  & P\left( A\cup B\cup C \right)+P\left( \bar{A}\cup \bar{B}\cup \bar{C} \right)=1 \\
 & \Rightarrow P\left( A\cup B\cup C \right)=1-P\left( \bar{A}\cup \bar{B}\cup \bar{C} \right) \\
\end{align}$
Here also events $A$ , $B$ and $C$ are independent events, then we will have
$P\left( A\cup B\cup C \right)=1-P\left( {\bar{A}} \right).P\left( {\bar{B}} \right).P\left( {\bar{C}} \right)$
Now we will use the probability of a complementary event as $P\left( {\bar{A}} \right)=1-P\left( A \right)$ in the above equation and substitute the known values. After simplifying the above equation we will get the same result as we get in our solution.