# 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

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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 $P\big(\frac{B}{G}\big)$

$\frac{P(B).P\big(\frac{G}{B}\big)}{P(A).P\big(\frac{G}{A}\big)+P(B).P\big(\frac{G}{B}\big)}$

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