Question 971466: A bag contains 9 red marbles and 5 yellow marbles. You are asked to draw 3 marbles from the bag without replacement. In how many ways can you draw 2 reds and 1 yellow?
I do not know how to solve this problem. Please provide me with an explanation to the solution and the answer.
Thank you!
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! prob 2 reds and 1 yellow Without replacement,the denominator changes, and so does the numerator if a given marble is chosen. There are 9 ways out of 14 to get a red first. Once a red has been chosen, there are 13 marbles left, but only 8 red ones.
(9/14)*(8/13)*(5/12)
There are 3 different ways this will occur, if the yellow is drawn 1st, 2nd, or third.
(5/14)(9/13)(8/12) is another way, but this is the same fraction when multiplied.
In all cases, however, the denominator is the same and the numerator will multiply out the same, even if it is over a different denominator.
(360/2184)*3=1080/2184
P= 0.495
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