Two marbles are drawn without replacement from a box with 3 white, 2 green, and 1 blue marble. Find the probability of drawing one white marble and one blue marble.

Question

anonymous · Answer

your help is greatly appreciated

kc_kennylau · Answer

First you draw one white marble

Then you'll be left with 2 white 2 green 1 blue

Then you draw one blue marble

anonymous · Answer

I think 4/15 is the answer. Can you confirm?

anonymous · Answer

First, I'm assuming that the order in which the marbles are drawn is not important. Thus, the probability of drawing a white marble and a blue marble is:

$$ = (\frac{ 3 }{ 6 })(\frac{ 1 }{ 5 }) + (\frac{ 1 }{ 6 })(\frac{ 3 }{ 5 }) = \frac{ 1 }{ 5 }$$

anonymous · Answer

I Think the answer is $$\frac{ 1 }{ 10 }$$ because the desired event is one white and one blue without replacement and the total number of balls is 6 then:
$$\frac{ 3 }{ 6 }  \frac{ 1 }{ 5 } = \frac{ 3 }{ 30 } = \frac{ 1 }{ 10 } $$

anonymous · Answer

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