There is a fair die with brown, black, and white colors on two of the six faces each being tossed continously (each color has two faces.) Y is the random variable having the value k if
(1) 2 of 3 colors appear in first k-1 tosses
(2) all colors appear in first k tosses.
What is E[Y]?
I think i should compute the probability of the opposite (not all colors appear in first k tosses.) There is a formula $$E[Y]=\sum_{k=0}^{infinity} P(Y>k)$$ i believe i can use. Since for k tosses, there are P(Y>k) chances that not all colors apepar in first k tosses. how can i apply this formula?
The probability that exactly one colour has appeared in the first $k$ throws is $\dfrac{{3 \choose 1}\cdot 1^k}{3^k}$ for $k \ge 1$
The probability that exactly two colours have appeared in the first $k$ throws is $\dfrac{{3 \choose 2}\cdot \left(2^k - {2 \choose 1}\cdot 1^k\right)}{3^k}$ for $k \ge 1$
So adding these together and simplifying: $P(Y>k) = \dfrac{2^k - 1}{3^{k-1}}$ for $k \ge 1$ and you can do your infinite sum using geometric series
Remember that $P(Y>0)=1$