Question

Three are three urns A B and C. Urn A contains 4 red and 4 green balls, Urn B contains 3 red and 5 green balls, Urn C contains 5 red and 2 green balls. One ball is drawn from each urns. What is probability that out of these three drawn balls one is red and two are green?

Answer 1

Dear student
$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{follwing}\mathrm{events}: \\ {\mathrm{E}}_1=\mathrm{Ball}\mathrm{drawn}\mathrm{from}\mathrm{urn}\mathrm{A}\mathrm{is}\mathrm{green} \\ {\mathrm{E}}_2=\mathrm{Ball}\mathrm{drawn}\mathrm{from}\mathrm{urn}\mathrm{B}\mathrm{is}\mathrm{green} \\ {\mathrm{E}}_3=\mathrm{Ball}\mathrm{drawn}\mathrm{from}\mathrm{urn}\mathrm{C}\mathrm{is}\mathrm{green} \\ \mathrm{Then} \\ \mathrm{P}\left({\mathrm{E}}_1\right)=\frac{4}{8}=\frac{1}{2},\mathrm{P}\left({\mathrm{E}}_2\right)=\frac{5}{8},\mathrm{P}\left({\mathrm{E}}_3\right)=\frac{2}{7} \\ \mathrm{Theerfore}, \\ \mathrm{P}\left({\overline{\mathrm{E}}}_1\right)=\mathrm{Ball}\mathrm{drawn}\mathrm{from}\mathrm{urn}\mathrm{A}\mathrm{is}\mathrm{red}=1-\mathrm{P}\left({\mathrm{E}}_1\right)=1-\frac{1}{2}=\frac{1}{2} \\ \mathrm{P}\left({\overline{\mathrm{E}}}_2\right)=\mathrm{Ball}\mathrm{drawn}\mathrm{from}\mathrm{urn}\mathrm{B}\mathrm{is}\mathrm{red}=1-\mathrm{P}\left({\mathrm{E}}_2\right)=1-\frac{5}{8}=\frac{3}{8} \\ \mathrm{P}\left({\overline{\mathrm{E}}}_3\right)=\mathrm{Ball}\mathrm{drawn}\mathrm{from}\mathrm{urn}\mathrm{C}\mathrm{is}\mathrm{red}=1-\mathrm{P}\left({\mathrm{E}}_3\right)=1-\frac{2}{7}=\frac{5}{7} \\ \mathrm{Now},\mathrm{two}\mathrm{green}\mathrm{balls}\mathrm{and}\mathrm{one}\mathrm{red}\mathrm{ball}\mathrm{can}\mathrm{be}\mathrm{drawn}\mathrm{in}\mathrm{the}\mathrm{following} \\ \mathrm{mutually}\mathrm{exclusive}\mathrm{ways}: \\ \left(\mathrm{I}\right)\mathrm{Green}\mathrm{from}\mathrm{urn}\mathrm{A},\mathrm{green}\mathrm{from}\mathrm{urn}\mathrm{B}\mathrm{and}\mathrm{red}\mathrm{from}\mathrm{urn}\mathrm{C}\mathrm{ie}.{\mathrm{E}}_1\cap {\mathrm{E}}_2\cap {\overline{\mathrm{E}}}_3 \\ \left(\mathrm{II}\right)\mathrm{Green}\mathrm{from}\mathrm{urn}\mathrm{A},\mathrm{red}\mathrm{from}\mathrm{urn}\mathrm{B}\mathrm{and}\mathrm{green}\mathrm{from}\mathrm{urn}\mathrm{C}\mathrm{ie}.{\mathrm{E}}_1\cap {\overline{\mathrm{E}}}_2\cap {\mathrm{E}}_3 \\ \left(\mathrm{III}\right)\mathrm{Red}\mathrm{from}\mathrm{urn}\mathrm{A},\mathrm{green}\mathrm{from}\mathrm{urn}\mathrm{B}\mathrm{and}\mathrm{green}\mathrm{from}\mathrm{urn}\mathrm{C}\mathrm{ie}.{\overline{\mathrm{E}}}_1\cap {\mathrm{E}}_2\cap {\mathrm{E}}_3 \\ \mathrm{Therefore}, \\ \mathrm{Required}\mathrm{probability} \\ =\mathrm{P}\left(\mathrm{I}\right)+\mathrm{P}\left(\mathrm{II}\right)+\mathrm{P}\left(\mathrm{III}\right) \\ =\mathrm{P}\left({\mathrm{E}}_1\cap {\mathrm{E}}_2\cap \overline{{\mathrm{E}}_3}\right)+\mathrm{P}\left({\mathrm{E}}_1\cap {\overline{\mathrm{E}}}_2\cap {\mathrm{E}}_3\right)+\mathrm{P}\left({\overline{\mathrm{E}}}_1\cap {\mathrm{E}}_2\cap {\mathrm{E}}_3\right) \\ =\mathrm{P}\left({\mathrm{E}}_1\right)\mathrm{P}\left({\mathrm{E}}_2\right)\mathrm{P}\left({\overline{\mathrm{E}}}_3\right)+\mathrm{P}\left({\mathrm{E}}_1\right)\mathrm{P}\left({\overline{\mathrm{E}}}_2\right)\mathrm{P}\left({\mathrm{E}}_3\right)+\mathrm{P}\left({\overline{\mathrm{E}}}_1\right)\mathrm{P}\left({\mathrm{E}}_2\right)\mathrm{P}\left({\mathrm{E}}_3\right) \\ =\frac{1}{2}\times \frac{5}{8}\times \frac{5}{7}+\frac{1}{2}\times \frac{3}{8}\times \frac{2}{7}+\frac{1}{2}\times \frac{5}{8}\times \frac{2}{7} \\ =\frac{25}{112}+\frac{6}{112}+\frac{10}{112} \\ =\frac{41}{112} \end{gathered}$$
Regards