Assume we have a 3 sided biased die. The probability that it lands on 1: $P(1) = \frac{2}{10} $. The probability that it lands on 2: $P(2) = \frac{3}{10}$, and $P(3) = \frac{5}{10}$.
Now we roll the die until we see the first 1. And Y is the number of total die tosses. Now I believe we can think of Y as a geometric random variable (please correct me if I am wrong).
The question is if Y=7, What is the expected number of 3's?
Now we know the 7th toss was a 1 so I believe we only need to check the first 6th tosses. This is where I get a bit confused. We also know the first 6 tosses were not 1. So I only expect to see 2 or 3. Then I believe we only need to consider the possibility of getting 3's knowing that they were not 1. Am I correct? So we can do Bayes:
$$P(3 | \neq1) = \frac{P(3\ \& \neq 1) }{P(\neq1)} = \frac{P(3)}{P(\neq1)} = \frac{5/10}{8/10} = \frac{5}{8}$$ So the expected number of 3's become: $6$ tosses x 5/8 probability = 30/8 $\approx$ 3.75. So my assumption is the expected number of 3's is 3.75. Are my assumptions correct? I am confused about the first 6 tosses.