SOLUTION: An urn contains four red balls, two green balls, and two yellow balls. Three balls will be drawn from the urn, one at a time, at random. If the balls are drawn without replaceme

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Question 1185572: An urn contains four red balls, two green balls, and two yellow balls. Three balls will be drawn from the urn, one at a time, at random.
If the balls are drawn without replacement (i.e. when a ball is drawn it is not placed back into the urn before the next draw), what is the probability the first is green, the second is yellow, and the third is red?

If the balls are drawn with replacement (i.e. when a ball is drawn it is placed back into the urn before the next draw), what is the probability the first is green, the second is yellow, and the third is red?
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Enter your answers as whole numbers or fractions in lowest terms.
Answer by math_helper(2461)   (Show Source): You can put this solution on YOUR website!


4 R

2 G

2 Y

---

8 total 



Without replacement: 



P(1st is G, 2nd is Y, 3rd is R) = (2/8)(2/7)(4/6) = 16/336 = 1/21



The 2/7 comes from the fact that only 7 remain after the first draw, and there are 2 Yellow so 2/7 represents the probability of a favorable 2nd draw outcome.  Similar for 3rd factor relating to the draw of Red.



With replacement:



In this case, the denominator stays constant at 8, as we are replacing after each selection:  P(G, Y, R) = (2/8)(2/8)(4/8) = 16/512 = 1/32

Qualitatively, this makes sense: say we pick G on 1st draw.  Since we are replacing that G into the urn, it slightly reduces the probability of selecting a Y on the 2nd draw.  Similar for R on 3rd draw.

 

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