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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\[(i)\;Bag \;2\quad(ii)\;Bag\; 3\]

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- \(E_1\)=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(\(E_3/B\))=\(\Large\frac{\frac{2}{6}\times1}{{\frac{1}{6}\times0+\frac{2}{6}\times\frac{1}{3}+\frac{3}{6}\times1}}\)

=\(\Large\frac{\frac{3}{6}}{\frac{11}{18}}\)

=\(\Large\frac{9}{11}\)

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