# 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