Container A holds 4 red balls and 6 green balls; containers B and C each hold 6 red balls and 3 green balls. A container is selected at random and then a ball is randomly selected from that container. What is the probability that the ball selected is green? Express your answer as a common fraction.
+128407 Container A holds 4 red balls and 6 green balls; containers B and C each hold 6 red balls and 3 green balls. A container is selected at random and then a ball is randomly selected from that container. What is the probability that the ball selected is green? Express your answer as a common fraction.
Here's my best attempt :
P(A is selected and a green ball is selected form A) = (1/3)(6/10) = (1/3) (3/5) = 1/5
P( B is selected and a green ball is selected from B )= (1/3) (3/9) = 1/9
P( C is seclected and a green ball is seected form C) = (1/3) (3/9) = 1/9
Total probability = 1 / 5 + 2 ( 1 / 9 ) = 1/5 + 2/9 = [ 9 + 10 ] / 45 = 19 / 45
