There are three bags B 1 ,   B 2 and B 3 . The bag B 1 contains 5 red and 5 green balls, B 2 contains 3 red and 5 green balls, and B 3 contains 5 red and 3 green balls, Bags B 1 ,   B 2 and B 3 have probabilities 3 10 , 3 10 and 4 10 respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?
C. Probability that the chosen ball is green, given that the selected bag is B 3 , equals 3 8
Let us define following events B 1 : Event of selection of bag 1 ( 5 Red, 5 Green) B 2   : Event of selection of bag 2 ( 3 Red, 5 Green) B 3 : Event of selection of bag 3 ( 5 Red, 3 Green) R = Event of selection of Red ball G = Event of selection of Green ball Option ( C ) : P G B 3 = 3 8 (as bag 3 has been selected now total balls in bag 3 = 8 , green ball = 3 ) Option ( B ) : P G = P B 1 P G B 1 + P B 2 P G B 2 + P B 3 P G B 3 = 3 10 × 5 10 + 3 10 × 5 8 + 4 10 × 3 8 = 39 80 Option ( D ) : P B 3 G = P B 3 ∩ G P G = P B 3 P G B 3 P ( G ) = 4 10 × 3 8 35 80   ∴   P G = 39 80 = 4 13 Option ( A ) : P B 3 ∩ G = P B 3 P G B 3 P B 3 ∩ G = P B 3 P G B 3 = 3 20
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