Question

The probability that or student A can solve a question is 6/7 and that of another student B solving the question is 3/4. Assuming that the 2 events "A can solve the question" and "B can solve the question " are independent, find the probability that only one of them solves the question.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{E}\mathrm{and}\mathrm{F}\mathrm{be}\mathrm{the}\mathrm{events}\mathrm{that}\mathrm{A}\mathrm{solves}\mathrm{the}\mathrm{question}\mathrm{and}\mathrm{B}\mathrm{solves}\mathrm{the}\mathrm{question}\mathrm{respectively}. \\ \mathrm{Now},\mathrm{P}\left(\mathrm{E}\right)=\frac{6}{7}\Rightarrow \mathrm{P}\left(\mathrm{A}\mathrm{does}\mathrm{not}\mathrm{solve}\mathrm{the}\mathrm{question}\right)=\mathrm{P}\left(\overline{\mathrm{E}}\right)=1-\frac{6}{7}=\frac{1}{7} \\ \mathrm{P}\left(\mathrm{F}\right)=\frac{3}{4}\Rightarrow \mathrm{P}\left(\mathrm{B}\mathrm{does}\mathrm{not}\mathrm{solve}\mathrm{the}\mathrm{question}\right)=\mathrm{P}\left(\overline{\mathrm{F}}\right)=1-\frac{3}{4}=\frac{1}{4} \\ \mathrm{Required}\mathrm{probability}=\mathrm{P}\left(\overline{\mathrm{E}}\mathrm{F}\right)+\mathrm{P}\left(\mathrm{E}\overline{\mathrm{F}}\right) \\ =\mathrm{P}\left(\overline{\mathrm{E}}\right).\mathrm{P}\left(\mathrm{F}\right)+\mathrm{P}\left(\mathrm{E}\right).\mathrm{P}\left(\overline{\mathrm{F}}\right) \\ =\frac{1}{7}\times \frac{3}{4}+\frac{6}{7}\times \frac{1}{4} \\ =\frac{3}{28}+\frac{6}{28} \\ =\frac{9}{28} \end{gathered}$$