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?
Algebra ->
Probability-and-statistics
-> 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?
Log On
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
--------
P(green/black pair) = 18/45 = 2/5
------------
Cheers,
Stan H.
------------------
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%.
(