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# Three bags contain a number of red and white balls as follows:

Bag 1:3 red balls,Bag 2:2 red balls and 1 white ball. Bag 3:3 white balls. The probability that bag i will be chosen and a ball is selected from it is $\Large \frac{i}{6}$, i=1,2,3.If a white ball is selected,what is th probability that it came from$(i)\;Bag \;2\quad(ii)\;Bag\; 3$

Toolbox:
• $E_1$=selecting $1^{st}$ bag
• $E_1$=selecting$2^{nd}$ bag
• $E_1$=selecting $^3{rd}$ bag
• B= $getting\; a\;white\; ball$
• 1=P(E$_2$/B)=$\Large\frac{{P(E_2)}{P(B/E_2)}}{{P(E_1)}{P(B/E_1)}+{P(E_2)}{P(B/E_2)}+{P(E_3)}{p(B/E_3)}}$
• 2=P(E$_3$/B)=$\Large\frac{{P(E_3)}{P(B/E_2)}}{{P(E_1)}{P(B/E_1)}+{P(E_2)}{P(B/E_2)}+{P(E_3)}{p(B/E_3)}}$
• 3=P(E$_3$/B)=$\Large\frac{{P(E_3)}{P(B/E_2)}}{{P(E_1)}{P(B/E_1)}+{P(E_2)}{P(B/E_2)}+{P(E_3)}{p(B/E_3)}}$
1=P($E_2/B$)=$\Large\frac{\frac{2}{6}\times\frac{1}{3}}{{\frac{1}{6}\times0+\frac{2}{6}\times\frac{1}{3}+\frac{3}{6}\times1}}$
=$\Large\frac{\frac{2}{18}}{\frac{1}{9}+\frac{3}{6}}$
=$\Large\frac{2}{11}$
2=P($E_3/B$)=$\Large\frac{\frac{2}{6}\times1}{{\frac{1}{6}\times0+\frac{2}{6}\times\frac{1}{3}+\frac{3}{6}\times1}}$
=$\Large\frac{\frac{3}{6}}{\frac{11}{18}}$
=$\Large\frac{9}{11}$

edited Jun 4, 2013