Assume that the box contains 6 balls: 1 white, 2 yellow, and 3 green. Balls are drawn in succession without replacement, and their colors are noted until a white ball is drawn. How many outcomes are there in the sample space?

Question

pulsified333 · Answer

I thought there should be 33 in the space but its wrong

anonymous · Answer

12C3 = 12! / (9! 3!) = 12 * 11 * 10 / (3 * 2 * 1) = 220 different combinations of three balls among the 12 balls in the box. This is the number of different possible 3-ball draws. 

There are 
6C3 = 6! (3! 3!) = 20 combinations of 3 balls that are all red. 

There are 
4C3 = 4 combinations of 3 balls that are all blue. 

There are no combinations of 3 balls that are all green. 

So the probability of drawing 3 balls the same color, without replacement, is 
(20 + 4) / 220 = 24/220 = 6/55 = about 10.9%

pulsified333 · Answer

there are no red balls in this problem

anonymous · Answer

?? Let me re-do. This was my problem, I've had this before.

pulsified333 · Answer

ok

anonymous · Answer

I did this awhile ago, mine was just with red instead of white, so I'm not sure.

dan815 · Answer

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