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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.

 Mar 25, 2021
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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

 

cool cool cool

 Mar 25, 2021

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