The Probability of A, B, C solving a problem are \[\dfrac{1}{3},\dfrac{2}{7},\dfrac{3}{8}\] respectively. If all three solve the problem simultaneously, the probability that exactly one of them will solve it.
A. \[\dfrac{{25}}{{168}}\]
B. \[\dfrac{{25}}{{56}}\]
C. \[\dfrac{{20}}{{168}}\]
D. \[\dfrac{{30}}{{168}}\]

Hint: In this problem, we have to find the probabilities such that when one of them is able to solve the remaining must fail in this way we will need to make 3 separate cases to solve each one of them and then add them all to get the final answer.

Complete step-by-step answer:
Case I:
Probability of A solving (remaining fail)
This will be given by \[P(A) \times P\left( {{B^C}} \right) \times P\left( {{C^C}} \right)\]
Now we know that
\[\begin{array}{l}
P(A) = \dfrac{1}{3}\\
P\left( {{B^C}} \right) = \left( {1 - \dfrac{2}{7}} \right) = \dfrac{{7 - 2}}{7} = \dfrac{5}{7}\\
P\left( {{C^C}} \right) = \left( {1 - \dfrac{3}{8}} \right) = \dfrac{{8 - 3}}{8} = \dfrac{5}{8}
\end{array}\]
Which means that
\[P(A) \times P\left( {{B^C}} \right) \times P\left( {{C^C}} \right) = \dfrac{1}{3} \times \dfrac{5}{7} \times \dfrac{5}{8} = \dfrac{{25}}{{168}}\]
Case II:
Probability of B solving (remaining fail)
This will be given by \[P\left( {{A^C}} \right) \times P\left( B \right) \times P\left( {{C^C}} \right)\]
Now we know that
\[\begin{array}{l}
P\left( {{A^C}} \right) = \left( {1 - \dfrac{1}{3}} \right) = \dfrac{{3 - 1}}{3} = \dfrac{2}{3}\\
P\left( B \right) = \dfrac{2}{7}\\
P\left( {{C^C}} \right) = \left( {1 - \dfrac{3}{8}} \right) = \dfrac{{8 - 3}}{8} = \dfrac{5}{8}\\
\therefore P\left( {{A^C}} \right) \times P\left( B \right) \times P\left( {{C^C}} \right) = \dfrac{2}{3} \times \dfrac{2}{7} \times \dfrac{5}{8} = \dfrac{{20}}{{168}}
\end{array}\]
Case III:
Probability of C solving (remaining fail)
This will be given by \[P\left( {{A^C}} \right) \times P\left( {{B^C}} \right) \times P\left( C \right)\]
Now we know that
\[\begin{array}{l}
P\left( {{A^C}} \right) = \left( {1 - \dfrac{1}{3}} \right) = \dfrac{{3 - 1}}{3} = \dfrac{2}{3}\\
P\left( {{B^C}} \right) = \left( {1 - \dfrac{2}{7}} \right) = \dfrac{{7 - 2}}{7} = \dfrac{5}{7}\\
P\left( C \right) = \dfrac{3}{8}\\
\therefore P\left( {{A^C}} \right) \times P\left( {{B^C}} \right) \times P\left( C \right) = \dfrac{2}{3} \times \dfrac{5}{7} \times \dfrac{3}{8} = \dfrac{{30}}{{168}}
\end{array}\]

So, the correct answer is “Option D”.

Note: Here Students often get confused about when to add and when to multiply. They often multiply in between the cases and add all the probability individually which is a wrong concept.