Given : let 
Now, 
![\begin{array}{l} =\mathrm{P}(\mathrm{A} \cap \bar{B} \cap \bar{C})+\mathrm{P}(\mathrm{B} \cap \bar{A} \cap \bar{C})+\mathrm{P}(\mathrm{C} \cap \bar{A} \cap \bar{B}) \\ =\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\bar{B}) \times \mathrm{P}(\bar{C})+\mathrm{P}(\mathrm{B}) \times \mathrm{P}(\bar{A}) \times \mathrm{P}(\bar{C})+\mathrm{P}(\mathrm{C}) \times \mathrm{P}(\bar{B}) \times \mathrm{P}(\bar{A}) \\ =\left[\frac{1}{3} \times \frac{3}{4} \times \frac{5}{6}\right]+\left[\frac{1}{4} \times \frac{2}{3} \times \frac{5}{6}\right]+\left[\frac{1}{6} \times \frac{3}{4} \times \frac{2}{3}\right] \\ =\frac{15}{72}+\frac{10}{72}+\frac{6}{72} \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-73e25e2050ecf76f9ef19b972779bd59_l3.png "Rendered by QuickLaTeX.com")
Therefore, The probability that excatly one of them will solve the problem is 
The probabilities of 
