Question 1075427: A bag consists 3 red marbles, 3 blue marbles, 3 green marbles and 2 yellow marbles, all of which are distinguishable from each other. Four marbles are drawn from the bag (without replacement). 1. How many possible draws are there? 2. How many draws contain 4 difference colored marbles? 3. How many draws contain at least 2 red marbles? 4. How many draws contain at least one green marble and no red marble?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! with all 11 distinguishable from one another, there are 11C4 or 330 different ways to draw them.
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draws that contain one of each are
3C1*3C1*3C1*2C1=54
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2 red marbles can occur in 6 possible ways in drawing 4 (4C2) plus 4 ways in 4C3, so 10 ways total.
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At least one green can occur 4 ways for 1, 6 ways for 2, and 4 ways for 3. If one green is chosen, there are 3 other marbles left out of 5 that can be chosen or 5C3=10. There are 10 ways.
For 2 green, there are now 2 other marbles out of 5 allowed or 10 ways.
For 3 green, there is now one other choice out of 5 or 5 ways.
There are 25 ways to choose at least 1 green and no red.
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