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A bag A contains 9 black balls, 6 white balls and 3 red balls. A bag B contains 6 black balls, 2 white balls and 4 green balls. Ali takes out 1 ball from each bag randomly. When Ali takes out 1 ball from one bag, he will put it into the other bag and then takes out one ball from that bag. Find the probability that

(a) the ball is black from bag A, followed by white from bag B,
(b) both the balls are white in colour,
(c) the ball is black or white from bag B, followed by red from bag A,
(d) both the balls are of different colours,
(e) both the balls are not black or white in colours.

Solution:

(a) $\displaystyle\frac{9}{18}\times\frac{2}{13}=\frac{1}{13}$

(b) Probability of white ball from bag A, followed by white ball from bag B= $\displaystyle=\frac{1}{2}\times\frac{6}{18}\times\frac{3}{13}=\frac{1}{26}$

Probability of white from B, followed by white from A= $\displaystyle=\frac{1}{2}\times\frac{2}{12}\times\frac{7}{19}=\frac{7}{228}$

Total prob= $\displaystyle\frac{205}{2964}$

$\displaystyle\frac{8}{12}\times\frac{3}{19}=\frac{2}{19}$

(d) Prob of both red = P(red from A, followed by red from B)= $\displaystyle\frac{1}{2}\times\frac{3}{18}\times\frac{1}{13}=\frac{1}{156}$

P(both green)=P(green from B, followed by green from A)= $\displaystyle\frac{1}{2}\times\frac{4}{12}\times\frac{1}{19}=\frac{1}{114}$

P(both black)=P(black from A, followed by black from B)+P(black from B, followed by black from A)= $\displaystyle\frac{1}{2}\times\frac{9}{18}\times\frac{7}{13}+\frac{1}{2}\times\frac{6}{12}\times\frac{10}{19}=\frac{263}{988}$

P(both white)= $\displaystyle\frac{205}{2964}$ (from part b)

$\displaystyle 1-\frac{1}{156}-\frac{1}{114}-\frac{263}{988}-\frac{205}{2964}=\frac{1925}{2964}$

(e)

P(neither black nor white from A, followed by neither black nor white from B)= $\displaystyle\frac{1}{2}\times\frac{3}{18}\times\frac{5}{13}=\frac{5}{156}$

P(neither black nor white from B, followed by neither black nor white from A)= $\displaystyle\frac{1}{2}\times\frac{4}{12}\times\frac{4}{19}=\frac{2}{57}$