if odds against solving a question by three students are 2:1, 5:2 ad 5:3 respectively the probability that question is solved by only one student

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$$\begin{gathered} \mathrm{Let} \mathrm{Student} \mathrm{be} \mathrm{A}, \mathrm{B} \mathrm{and} \mathrm{C} \\ \mathrm{Odd} \mathrm{against} \mathrm{A}=\frac{2}{1} \\ \Rightarrow \mathrm{Probability} \mathrm{of} \mathrm{A} \mathrm{solving} \mathrm{question}=\frac{1}{1+2}=\frac{1}{3} \\ \mathrm{Odd} \mathrm{against} \mathrm{B}=\frac{5}{2} \\ \Rightarrow \mathrm{Probability} \mathrm{of} \mathrm{B} \mathrm{solving} \mathrm{question}=\frac{2}{5+2}=\frac{2}{7} \\ \mathrm{Odd} \mathrm{against} \mathrm{C}=\frac{5}{3} \\ \Rightarrow \mathrm{Probability} \mathrm{of} \mathrm{A} \mathrm{solving} \mathrm{question}=\frac{3}{5+3}=\frac{3}{8} \\ \mathrm{Probabilty} \mathrm{that} \mathrm{only} \mathrm{one} \mathrm{student} \mathrm{solves} \\ =\left(\mathrm{A} \mathrm{solves} \mathrm{and} \mathrm{B} \mathrm{and} \mathrm{C} \mathrm{don}\text{'}\mathrm{t}\right) \mathrm{or} \left(\mathrm{B} \mathrm{solves} \mathrm{and} \mathrm{C} \mathrm{and} \mathrm{A} \mathrm{don}\text{'}\mathrm{t}\right) \mathrm{or} \left(\mathrm{C} \mathrm{solves} \mathrm{and} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{don}\text{'}\mathrm{t}\right) \\ =\frac{1}{3}\left(1-\frac{2}{7}\right)\left(1-\frac{3}{8}\right)+\frac{2}{7}\left(1-\frac{3}{8}\right)\left(1-\frac{1}{3}\right)+\frac{3}{8}\left(1-\frac{1}{3}\right)\left(1-\frac{2}{7}\right) \\ =\frac{25}{168}+\frac{20}{168}+\frac{30}{168} \\ =\frac{25+20+30}{168} \\ =\frac{75}{168}=\frac{25}{56}\left(\mathrm{Answer}\right) \end{gathered}$$
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