I m asking whether sum can be done in any other way . u have done in sameway in which i have given

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{The} \mathrm{way} \mathrm{it} \mathrm{is} \mathrm{solved} \mathrm{in} \mathrm{the} \mathrm{book} \mathrm{is} \mathrm{the} \mathrm{smartest} \mathrm{and} \mathrm{shortest} \mathrm{way} \mathrm{to} \mathrm{solve} \mathrm{this} \\ \mathrm{problem}. \mathrm{Now} \mathrm{let} \mathrm{us} \mathrm{understand} \mathrm{other} \mathrm{method}. \\ \mathrm{Required} \mathrm{Probabilty} \\ =\left(\mathrm{A} \mathrm{solves} \mathrm{but} \mathrm{B} \mathrm{and} \mathrm{C} \mathrm{don}\text{'}\mathrm{t} \mathrm{solve}\right) \mathrm{OR} \left(\mathrm{B} \mathrm{solves} \mathrm{but} \mathrm{C} \mathrm{and} \mathrm{A} \mathrm{don}\text{'}\mathrm{t} \mathrm{solve}\right) \mathrm{OR} \\ \left(\mathrm{C} \mathrm{solves} \mathrm{but} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{don}\text{'}\mathrm{t} \mathrm{solve}\right) \mathrm{OR} \left(\mathrm{A} \mathrm{and} \mathrm{B} \mathrm{solve} \mathrm{but} \mathrm{C} \mathrm{does} \mathrm{not} \mathrm{solve} \right) \mathrm{OR} \\ \left(\mathrm{B} \mathrm{and} \mathrm{C} \mathrm{solve} \mathrm{but} \mathrm{A} \mathrm{does} \mathrm{not} \mathrm{solve} \right) \mathrm{OR} \left(\mathrm{C} \mathrm{and} \mathrm{A} \mathrm{solve} \mathrm{but} \mathrm{B} \mathrm{does} \mathrm{not} \mathrm{solve} \right) \mathrm{OR} \\ \left(\mathrm{A} \mathrm{and} \mathrm{B} \mathrm{and} \mathrm{C} \mathrm{solve}\right) \\ =\mathrm{P}\left(\mathrm{A}\right)\left\{1-\mathrm{P}\left(\mathrm{B}\right)\right\}\left\{1-\mathrm{P}\left(\mathrm{C}\right)\right\}+\mathrm{P}\left(\mathrm{B}\right)\left\{1-\mathrm{P}\left(\mathrm{C}\right)\right\}\left\{1-\mathrm{P}\left(\mathrm{A}\right)\right\}+\mathrm{P}\left(\mathrm{C}\right)\left\{1-\mathrm{P}\left(\mathrm{A}\right)\right\}\left\{1-\mathrm{P}\left(\mathrm{B}\right)\right\} \\ +\mathrm{P}\left(\mathrm{A}\right) \mathrm{P}\left(\mathrm{B}\right) \left\{1-\mathrm{P}\left(\mathrm{C}\right)\right\}+\mathrm{P}\left(\mathrm{B}\right) \mathrm{P}\left(\mathrm{C}\right) \left\{1-\mathrm{P}\left(\mathrm{A}\right)\right\}+\mathrm{P}\left(\mathrm{C}\right) \mathrm{P}\left(\mathrm{A}\right) \left\{1-\mathrm{P}\left(\mathrm{B}\right)\right\}+\mathrm{P}\left(\mathrm{A}\right) \mathrm{P}\left(\mathrm{B}\right) \mathrm{P}\left(\mathrm{C}\right) \\ =\frac{1}{2}\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)+\frac{1}{3}\left(1-\frac{1}{4}\right)\left(1-\frac{1}{2}\right)+\frac{1}{4}\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \\ +\frac{1}{2}\times \frac{1}{3}\times \left(1-\frac{1}{4}\right)+\frac{1}{3}\times \frac{1}{4}\times \left(1-\frac{1}{2}\right)+\frac{1}{4}\times \frac{1}{2}\times \left(1-\frac{1}{3}\right)+\frac{1}{2}\times \frac{1}{3}\times \frac{1}{4} \\ =\frac{6}{24}+\frac{3}{24}+\frac{2}{24}+\frac{3}{24}+\frac{1}{24}+\frac{2}{24}+\frac{1}{24} \\ =\frac{6+3+2+3+1+2+1}{24} \\ =\frac{18}{24} \\ \mathbf{=}\frac{\mathbf{3}}{\mathbf{4}}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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