What is the expected number of marked fishes after 7 times? Given 10 fishes, each time one fish is chosen randomly, marked and returned to the pool. If a fish is already marked, it constitutes as a turn, and returned the the pool as well.
What is the expected number of marked fishes after 7 times?
Is it:
For each fish,
P(marked by 1st turn) OR  P(marked by 2nd turn)... OR P(marked by 7th turn) = $\frac{1}{10} * 7$
Hence, for ten fishes, the expected number of marked fishes after 7 times (by linearity): $$\frac{7}{10}*10=7.$$

This is a similar question from Brilliant.org where it asked about:
What is the expected number of unmarked fishes after 7 times?
The thought process would be for each time to not be marked: $\frac{9}{10}$.
So, P(unmarked in 1st) AND P(unmarked in 2nd) ... AND P(unmarked in 7th) = $\frac{9}{10}^7.$ Then it would be
$$\frac{9}{10}^7*10$$
Part of what I would like to clarify is whether my way of thinking is correct between AND and OR; not AND: multiply and OR: sum, but rather, did I construct the solution correctly for the marked version? My initial immediate thought for the marked fishes was to use $0.1^7$ instead of $\frac{1}{10} * 7$.
Sorry if what I am asking is unclear.

Just in case you do not want to do analytical math, we can use numeric simulation in R to answer this approximately.
If I name your ten fish using values from 1 to 10, each time we catch only one, mark and return it to the pool. That is sampling with replacement.
So we sample the pool 7 times, count the number of unique values.
    length(unique(sample(10,7, replace=T)))

We can simulate this process 100000 times to get the probability.
set.seed(1)
count <- vector()
for (i in 1:100000){
  count[i] <- length(unique(sample(10,7, replace=T)))
}
mean(count)

The result is about 5.2.
