Picking color randomly I have $7$ colors (white, black, green, orange, yellow, blue and red). I pick $3$ of them at random, and whichever color is drawn already is eliminated from the pool. What is the probability to draw each color?
I know for the first color, say red, the probability is $\frac 1 7$, correct?  But once it has been eliminated, what is the probability of drawing each other color for the next two draws?
 A: Let choose black as our favorite colour.
In the first round, you have a $\frac{1}{7}$ chance of picking black, in the second round, you have a $\frac{1}{6}$ chance of picking black, but only if you have not picked black already, which is a $\frac{6}{7}$ chance, so the chance of getting black in the second round is $\frac{6}{7} \times \frac{1}{6} = \frac{1}{7}$. In the last round, you have a $\frac{6}{7} \times \frac{5}{6} \times \frac{1}{5} =\frac{1}{7}$ chance of picking black, since you must pick a different colour in the first two rounds, and black in the last round.
So the chance of picking black is $\frac{1}{7} + \frac{1}{7} + \frac{1}{7} = \frac{3}{7}$, as is the chance of any other colour.
A: I wonder why all this computation is necessary. Suppose we transform the problem to: 
7 candidates are interviewed and 3 are to be selected
What is the probability that you are selected ?
Just $\dfrac37$, that's all !
A: If we take your given set of seven colors, and we want to draw red, and we draw three colors...
First, there is as you say, a probability of drawing red equal to $\frac17$.
If at first we do not draw red, there are six colors left. Not drawing red first is $\frac67$ and subsequently drawing red is $\frac16$. Multiply those to get $\frac6{42}=\frac17$.
If we don't draw red twice, which only happens with probability $\frac67×\frac56=\frac{30}{42}=\frac57$, then we have a remaining probability of $\frac15$. Multiplied together, that again gives $\frac17$.
Since these three classes of outcomes are disjoint, we can add their probabilities directly. The scenario you describe happens exactly $\frac37$ of the time.
That's a long way to go around, eh?
Let's put this problem another way: You draw all the colors. What's the probability that the red is in the first three draws? You can draw red in seven different ways: first, second, third, fourth, fifth, sixth, and seventh. Only three of those are first, second, or third. Ergo, $\frac37$.
