A problem in mathematics is given to three students A, B, C and their respective chance of solving the problem are $\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4}$. The probability that the problem will be solved is
A. $\dfrac{3}{4}$
B. $\dfrac{1}{2}$
C. $\dfrac{2}{3}$
D. $\dfrac{1}{3}$

Hint: Find the probability of A, B, and C not solving by using the fact that the sum of the probability of an event and the probability of the complement of the event is one. The probability that the problem will be solved is the same as the probability that at least one of A, B, and C solves the question.

Formula Used: $P\left( {\overline X } \right) = 1 - P\left( X \right)$

Complete step by step Solution:
It is given to us that,
The probability of A solving the question, $P\left( A \right) = \dfrac{1}{2}$
The probability of B solving the question, $P\left( B \right) = \dfrac{1}{3}$
The probability of C solving the question, $P\left( C \right) = \dfrac{1}{4}$
Therefore,
The probability of A not solving, $P\left( {\overline A } \right) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$
The probability of B not solving, $P\left( {\overline B } \right) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$
The probability of C not solving, $P\left( {\overline C } \right) = 1 - \dfrac{1}{4} = \dfrac{3}{4}$
The probability that the problem will be solved is the same as the probability that at least one of the three students solves the question.
The complement of the event where at least one of the three students solving the question is the event where none of the students can solve it = \[P\left( {\overline A } \right).P\left( {\overline B } \right).P\left( {\overline C } \right)\]
Therefore, the probability that problem will be solved is $1 - P\left( {\overline A } \right).P\left( {\overline B } \right).P\left( {\overline C } \right) = 1 - \dfrac{1}{4} = \dfrac{3}{4}$.

Therefore, the correct option is (A).

Note: Probability that A solves = $\dfrac{1}{2}$
Probability that A doesn’t solve, and B solves = $\dfrac{1}{2}.\dfrac{1}{3} = \dfrac{1}{6}$
Probability that A doesn’t solve, B doesn’t solve, and C solves = $\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{1}{4} = \dfrac{1}{{12}}$
Probability that at least one of them solves = $\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{{12}} = \dfrac{3}{4}$