SOLUTION: A bag contains 3 green balls,6 black balls and one white ball.if two balls are drawn at random without replacement,what is the probability of drawing a green and then a black ball?

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Question 1117838: A bag contains 3 green balls,6 black balls and one white ball.if two balls are drawn at random without replacement,what is the probability of drawing a green and then a black ball?
Found 2 solutions by stanbon, ikleyn:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
A bag contains 3 green balls,6 black balls and one white ball.if two balls are drawn at random without replacement,what is the probability of drawing a green and then a black ball?
# of green/black pairs:: 3*6 = 18
# of random pairs:: 10C2 = 5*9 = 45
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P(green/black pair) = 18/45 = 2/5
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Cheers,
Stan H.
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Answer by ikleyn(52800)   (Show Source): You can put this solution on YOUR website!
.
First, since the problem talks about the sequence of the balls ("a green and then a black ball"), it is clear that

the PERMUTATIONS, not combinations, work.


Next, the total number of the balls is 3 + 6 + 1 = 10.


Hence, the probability to get a green ball first is  .


Then the probability to get a black ball next is   = .


Thus the final answer is   =  =  = 0.2 = 20%.


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