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A bag A contains 4 green and 3 red balls and bag B contains 4 red and 3 green balls. One bag is taken at random and a ball is drawn and noted to be green.The probability that it comes from bag B is

(a) \(\frac{2}{7}\) 

(b) \(\frac{2}{3}\) 

(c) \(\frac{3}{7}\) 

(d) \(\frac{1}{3}\)

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Answer: (c)  \(\frac{3}{7}\) 

P(Drawing a green ball from bag A) = \(P\big(\frac{G}{A}\big) =\frac{4}{7}\)

P(Drawing a green ball from bag B)  =  \(P\big(\frac{G}{B}\big) =\frac{3}{7}\)

∴  Required probability \( P\big(\frac{B}{G}\big)\)


\(\frac{\frac{1}{2}.\frac{3}{7}}{\frac{1}{2}.\frac{4}{7}+\frac{1}{2}.\frac{3}{7}}\)  = \(\frac{3/14}{4/14+3/14} = \frac{3/14}{7/14} = \frac{3}{7}\)

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