Container I holds 8 red bs and 4 green bs; containers II and III each hold 2 red bs and 4 green bs. A container is selected at r

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Container I holds 8 red bs and 4 green bs; containers II and III each hold 2 red bs and 4 green bs. A container is selected at r

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Container I holds 8 red b***s and 4 green b***s; containers II and III each hold 2 red b***s and 4 green b***s. 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.

Guest Apr 11, 2015

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The probability is (1/3)*(4/12)+(1/3)*(4/6)+(1/3)*(4/6) = 5/9

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Alan Apr 12, 2015

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Alan · Answer 1 · 2015-04-12T08:33:47

The probability is (1/3)*(4/12)+(1/3)*(4/6)+(1/3)*(4/6) = 5/9

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Container I holds 8 red bs and 4 green bs; containers II and III each hold 2 red bs and 4 green bs. A container is selected at r

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Container I holds 8 red b***s and 4 green b***s; containers II and III each hold 2 red b***s and 4 green b***s. A container is selected at r

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Container I holds 8 red bs and 4 green bs; containers II and III each hold 2 red bs and 4 green bs. A container is selected at r

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