Probability of 4 colors sequence with repetition (8 choices of color) Knowing that the sequence contains exactly two colors, what is the probability that one of the two colors will be repeated exactly three times?
A = The probability that one of the two colors will be repeated exactly three times
B = The probability the sequence contains exactly two colors
I know there are $8^4 = 4096$ possible sequences.
Select 2 colors among 8 $ {8 \choose 2} = 28$
choose 2 place among 4  ${4 \choose 2} = 6$
I'm not sure but I think $P(A|B) = 8/14$
Knowing that a color is repeated at least twice, what is the probability that it will be repeated exactly
three times?
A =The probability that a color is repeated at least twice
B = The probability that a color is repeated exactly three times
 A: Assume colors are C-1, C-2, ..., C-8.
The easy way:
Since you are allowed to assume that exactly two colors occur in the sequence, and since the probability of exactly three out of four is unaffected by which colors are involved, assume, Without Loss of Generality that only colors C-1, C-2 are involved.
There are $2^4$ sequences possible, minus the two that only involve one color.  So the denominator will be $14$.
Assume that there are 3 of C-1, 1 of C-2.  There are 4 possible positions for C-2.  Therefore, there are 4 ways this can happen.  Multiply by 2 re 3 of C-2, 1 of C-1.
Final Answer: $\frac{8}{14}.$

The hard way : involves Bayes Theorem.
$D =$ denominator 
$N =$ numerator.
First, compute the probability of there being exactly two colors in the sequence.
This will be
$$D = \frac{\binom{8}{2} \times 14}{8^4}.$$
Then compute the probability of there being exactly two colors in the sequence and one of the two colors occurs exactly 3 times.
$$N = \frac{\binom{8}{2} \times 8}{8^4}.$$
Then $$\frac{N}{D} = \frac{8}{14}.$$
A: Your first answer is correct. It is not important which two colors are present and you can start from the point that the sequence of four has two colors.
Number of arrangements where two chosen colors are present $ = \displaystyle {2^4 - 2} = 14$. We basically take out two arrangements where only one of the two colors is present.
Arrangements where one of them is present exactly $1$ time (and so other $3$ times) $ = \displaystyle 2 \cdot 4 = 8$
So probability $ = \displaystyle \frac{8}{14} = \frac{4}{7}$.
For the second question, for a given color to repeat at least two times in sequence of four, below are the possibilities.
$(i)$ it repeats four times: $1$ arrangement
$(ii)$ it repeats three times: $4 \cdot 7 = 28$ arrangements
$(iii)$ it repeats two times: $\displaystyle {4 \choose 2} \cdot 7 + {4 \choose 2} \cdot2 \cdot{7 \choose 2} = 294 $ arrangements
Explanation -
In $(ii)$, we choose three positions out of four and for the last there is choice of $7$ colors.
In $(iii)$, we choose two positions for the given color and for the other two positions, we either have $1$ of the $7$ colors or we have $2$ of the $7$ colors and also their positions can swap.
Now probability of $(ii)$ is $\displaystyle \frac{(ii)}{(i)+(ii)+(iii)} = \frac{28}{323}$
A: 
Knowing that the sequence contains exactly two colors, what is the probability that one of the two colors will be repeated exactly three times?

It is given that there are only two colours in this particular sequence. How many possible sequences of length 4 can be built from a choice of two colours?
How many of these contain exactly three of the same colour?

Knowing that a color is repeated at least twice, what is the probability that it will be repeated exactly three times?

The sequence is of length four, and you know that two of the choices are already particular colour, call it colour $c$. So you only need to think about what the other two choices in the sequence might be. How many possibilities are there for the last two choices, and how many of them contain colour $c$ exactly once (so that the total sequence contains $c$ exactly three times)?
A: Using the definition of conditional probability, it is known that
$$\begin{split}P(repeat\_3|2\_colors)&=\frac{P(repeat\_3,2\_colors)}{P(2\_colors)}\\
&=\frac{2{4 \choose 3}{8 \choose 2}}{\left[{4\choose 1}+{4\choose 2}+{4\choose 3}\right]\cdot {8\choose 2}}\\
&=\frac{4}{7}\end{split}$$
The numerator comes from choosing 2 colors and three spots as you have done. However since the number of spots is not half the total number of spots you have to multiply by 2. The bottom is the number of ways you can get 2 colors.
Your answer for the first part looks correct. For the second part, again use the definition of conditional probability. We get
$$\begin{split}P(repeat\_thrice|repeat\_atleasttwice)&=\frac{P(repeat\_thrice,repeat\_atleasttwice)}{P(repeat\_atleasttwice)}\\
&=\frac{P(repeat\_thrice)}{P(repeat\_atleasttwice)}\\
&=\frac{\frac{{4\choose 3}}{8^4}*7}{1-\frac{{4\choose 1}7^3}{8^4}-\frac{7^4}{8^4}}\\
&=\frac{7/1024}{323/4096}\\
&=\frac{28}{323}\end{split}$$
