question_answer

Three bags contain a number of red and white balls as follows Bag I: 3 red balls, Bag II: 2 red balls and 1 white ball and Bag III: 3 white balls. The probability that bag will be chosen and a ball is selected from it is \[\frac{i}{6},\] where i =1, 2, 3. (i) What is the probability that a red ball is selected? (ii) If a white ball is selected, then, what is the probability that it came from bag III?

Answer:

Given, in bag I: 3 red balls and 0 white ball, in bag II: 2 red balls and 1 white ball and in bag III: 0 red ball and 3 white balls. Let \[{{E}_{1}},\,\,{{E}_{2}}\] and \[{{E}_{3}}\] be the events that bag I, bag II and bag III is selected and a ball is chosen from it. Then, \[P\,({{E}_{1}})=\frac{1}{6},\] \[P\,({{E}_{2}})=\frac{2}{6}\] and \[P\,({{E}_{3}})=\frac{3}{6}\] (i) Let E be the event that a red ball is selected. Then, probability that a red ball is selected, \[P(E)=P({{E}_{1}})\cdot P(E\,/{{E}_{1}})+P({{E}_{2}})\cdot P(E\,/{{E}_{2}})\]                                         \[+P({{E}_{3}})\cdot P(E\,/{{E}_{3}})\] \[=\frac{1}{6}\cdot \frac{3}{3}+\frac{2}{6}\cdot \frac{2}{3}+\frac{3}{6}\cdot 0\]             \[=\frac{1}{6}+\frac{2}{9}+0=\frac{3+4}{18}=\frac{7}{18}\] (ii) Let F be the event that a white ball is selected. Then, probability that a white ball is selected, \[P(F)=P({{E}_{1}})\cdot P(F\,/{{E}_{1}})+P({{E}_{2}})\cdot P(F\,/{{E}_{2}})\]                                  \[+P({{E}_{3}})\cdot P(F\,/{{E}_{3}})\] \[=\frac{1}{6}\cdot 0+\frac{2}{6}\cdot \frac{1}{3}+\frac{3}{6}\cdot 1=\frac{1}{9}+\frac{3}{6}=\frac{2+9}{18}=\frac{11}{18}\] Now, probability that a white ball came from bag III, \[P({{E}_{3}}/F)=\frac{P({{E}_{3}})P(F/{{E}_{3}})}{P({{E}_{1}})\cdot P(F/{{E}_{1}})+P({{E}_{2}})\cdot P(F/{{E}_{2}})}\]                                         \[+P({{E}_{3}})\cdot P(F/{{E}_{3}})\]               [using Baye's theorem] \[=\frac{\frac{3}{6}\cdot 1}{P(F)}=\frac{\frac{3}{6}}{\frac{11}{18}}=\frac{3}{6}\times \frac{18}{11}=\frac{9}{11}\]