+479 A bag has 4 red marbles, 5 white marbles, and 6 blue marbles. Three marbles are drawn from the bag (without replacement). What is the probability that they are all the same color?
+324 Lol, I never gave the answer.
Here is the answer:
\((\frac {4}{15} \cdot \frac {3}{14} \cdot \frac {2}{13})+(\frac {5}{15}\cdot\frac {4}{14}\cdot\frac {3}{13}) + (\frac {6}{15}\cdot \frac {5}{14}\cdot \frac {4}{13})\)
Compute it using any calculator. Don't expect us to do that as well. (Next time, use the class message board.)
+479 I don't take AoPS classes this is just a problem. I listen to video though.
Also tyty, ill try to get the reasoning?
+324 Sorry, had to eat dinner.
... And how did you know I was refering to AoPS. Whatever, irrelevent.
Here is why I did that.
So, if we took the probability that the red is first, then it is 4/15. Because there is no replacement, we have to remove one value from both numerator and denominator: (4-1)/(15-1). One mpre time for the third: (4-2)/(15-2). Then we multiply them all together. Careful: if you subtract 1/1 instead of 1 from the numerator and 1 from the denominator, then you will get a wrong number.
This in mind, tell me how I did the other 2 parts: white and blue. Just reply to this thread.
+118608 I cannot fault your logic Naciema. That is because it is correct! Thank you for your answer. 
I think your method gives a solution of 34/455
I did it a different way and got a different answer the same answer.
I said that there are 4C3=4 ways to chose 3 red balls
There are 5C3=10 ways to chose 3 white
and 6C3=20 ways to choose 3 blue.
That is a total of 34 favourable outcomes
The sample space is 15C3 = 286 ways 455 ways (error corrected)
34/286 = 17/143 34/455
Just as Nacirema found
