Probability that three chips of different colours are selected from a stack with $4$ red, $4$ blue, and $4$ yellow chips Hi I hope someone could help me out. I am really bad at calculating probability when it comes to combinations and there is a question that has been bugging me.
There are 12 chips numbered 1....12 with 3 different colours alternating. e.g 1 is red, 2 is yellow, 3 is blue, 4 is red, etc. All of these chips are stacked randomly. 3 chips are taken from the top randomly. What is the probability that all 3 taken chips are of different colours?
Can anyone suggest a way to approach this method? I can't come up with an approach. All I know that there is a total of 12C3 ways. is it wrong to say that i can have a total of 6 different ways (6!)?
Thanks a lot!
 A: Method 1: There are $\binom{12}{3}$ ways to select three of the twelve chips.  To get three different colours, we must select one of the four blue chips, one of the four red chips, and one of the four yellow chips.  Hence, the desired probability is 
$$\frac{\dbinom{4}{1}\dbinom{4}{1}\dbinom{4}{1}}{\dbinom{12}{3}}$$
Method 2:  There are $12 \cdot 11 \cdot 10$ ways to make an ordered selection of three chips.  For the favorable cases, there are $12$ ways to pick the first chip, $8$ ways to pick a second chip that is of a different colour than the first, and $4$ ways to pick a third chip that is of a different colour than the first two if two different colour chips have already been selected.  Hence, the desired probability is 
$$\frac{12 \cdot 8 \cdot 4}{12 \cdot 11 \cdot 10}$$
Method 3:  We pick a chip.  Eight of the eleven chips that remain are of a different colour than the first, so the probability that the second chip is of a different colour than the first is $8/11$.  If two different colour chips have been selected, four of the remaining ten chips are of a different colour than those already selected, so the probability that the third chip is of a different colour than the first two is $4/10$.  Hence, the desired probability is 
$$1 \cdot \frac{8}{11} \cdot \frac{4}{10}$$   
A: First, ask yourself what the desired method looks like. You're picking three chips of different color and there are $3$ colors, so you have to get exactly one red chip, exactly one yellow chip, and exactly one blue chip. Getting three chips of different color is equivalent to picking one red chip out of the four red chips, one yellow chip out of each of the four yellow chips, and one blue chip out of each of the four blue chips.
Observe that your choice of each chip of a certain color doesn't affect your choices of the other chips of the other colors. That is, no matter which red chip I pick, I can pick any one of the yellow chips, and any one of the blue chips. This means I take the product out of all my possible choices to consider all three-chip possibilities. To give you an example, if I have $5$ shirts, $3$ pairs of pants, and $2$ pairs of shoes, I have $5\times 3\times 2=30$ possible outfits, since I can wear any shirt with any pair of pants and any pair of shoes.
Now apply this to your example: how many ways can I pick a red chip, how many ways can I pick a yellow chip, and how many ways can I pick a blue chip? Take the product of these, and you'll know how many ways you can pick three chips of different color.
