Find probabilities and probability mass function Suppose an urn contains three balls, once black, one white, and one green. Assume
that balls are repeatedly drawn, one at a time, with replacement, and let X denote
the number of draws until each colour appears at least once.
(a) Find the probabilities $P(X > n)$ for $n = 0, 1, 2,....$
(b) Find the probability mass function (pmf) of X  
For part (a) we can say $P(X>n) = P(A_n \cup B_n \cup C_n)$ where $A_n$ is the event that no black balls are drawn, $B_n$ is the event that no white balls are drawn and $C_n$ is the event that no green balls are drawn
So, $P(A_n \cup B_n \cup C_n) = P(A_n) + P(B_n) + P(C_n) - P(A_n \cap B_n) - P(A_n \cap C_n) - P(B_n \cap C_n) + P(A_n \cap B_n \cap C_n)$
How do I compute these probabilities ?
 A: The probability that one particular ball is never drawn in $n$ tries is:
$$P(A_n) = P(B_n)= P(C_n) = {(\tfrac 2 3)}^{n}$$
The probability that two particular balls are never drawn in $n$ tries is:
$$P(A_n\cap B_n) = P(B_n\cap C_n)= P(A_n\cap C_n) = {(\tfrac 1 3)}^{n}$$
The probability that three particular balls are never drawn in $n$ tries is:
$$P(A_n\cap B_n\cap C_n) = {(\tfrac 0 3)}^{n}$$
So $P(X>n) = \frac{\cdot 2^n - 1}{3^{n-1}}$
Then $P(X=n) = P(X>n-1)-P(X>n)$
A: The case $n\leq 2$ is trivial. Assume $n > 2$: Following your notation the probabaility that you don't draw a certain ball until $n$ is $(2/3)^n$. $P(A_n \cap B_n \cap C_n)=0$ because you have to draw some balls. $A_n \cap B_n$ is the event that you draw $n$ green balls in a row. Did that help?
A: Certainly $n\geq3$ and say we're interested in $\mathcal P(X=n)$. There are $3^n$ possible outcomes. Now, if we needed $n$ drawings to get the three colors, that means that with $n-1$ drawings we had only two colors, say, colors B and W. There are $2^{n-1}$ of drawing only B's and W's. We take out the cases all B's and all W's, so we have $2^{n-1}-2$ possible drawings that include B and W but not G. Say our $n$-th drawing is G and we've got the three colors. The same reasoning goes for excluding $W$ or excluding $B$.
Hence
$$\mathcal P(X=n) =3\frac{2^{n-1}-2}{3^n}$$
Now it should be easy to get $\mathcal P(X>n)$
