Question Number 112163 by bobhans last updated on 06/Sep/20
$$\mathrm{In}\:\mathrm{a}\:\mathrm{class}\:\mathrm{of}\:\mathrm{30}\:\mathrm{pupils}\:,\:\mathrm{12}\:\mathrm{are}\:\mathrm{girls}\:\mathrm{and}\: \\ $$$$\mathrm{2}\:\mathrm{of}\:\mathrm{them}\:\mathrm{are}\:\mathrm{short}−\mathrm{sighted}.\:\mathrm{Among}\: \\ $$$$\mathrm{18}\:\mathrm{boys}\:,\:\mathrm{6}\:\mathrm{are}\:\mathrm{short}−\mathrm{sighted}\:.\:\mathrm{If}\:\mathrm{a}\:\mathrm{pupil} \\ $$$$\mathrm{is}\:\mathrm{selected}\:\mathrm{at}\:\mathrm{random},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{pupil}\:\mathrm{chosen}\:\mathrm{will}\:\mathrm{be}\: \\ $$$$\begin{cases}{\left(\mathrm{a}\right)\:\mathrm{a}\:\mathrm{girls}}\\{\left(\mathrm{b}\right)\:\mathrm{short}−\mathrm{sighted}}\\{\left(\mathrm{c}\right)\:\mathrm{a}\:\mathrm{short}−\mathrm{sighted}\:\mathrm{boy}}\end{cases} \\ $$
Answered by MJS_new last updated on 06/Sep/20
$$\left({a}\right)\:\frac{\mathrm{12}}{\mathrm{30}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\left({b}\right)\:\frac{\mathrm{8}}{\mathrm{30}}=\frac{\mathrm{4}}{\mathrm{15}} \\ $$$$\left({c}\right)\:\frac{\mathrm{6}}{\mathrm{30}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$
Commented by bobhans last updated on 06/Sep/20
$$\mathrm{hihihi}….\mathrm{thank}\:\mathrm{you} \\ $$
Commented by MJS_new last updated on 06/Sep/20
...after this hard job I need a break ��