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.

Let $X_t$ denote the number of marked fish after $t$ rounds.  Clearly, given $n=10$ fishes in total, and $X_t$ fishes marked after round $t$, you catch an already marked fish with probability $X_t/n$ and an unmarked fish with probability $1-X_t/n$ such that the conditional expectation $X_{t+1}$ is
\begin{align}
E(X_{t+1}|X_t)
  &=\frac {X_t} n \cdot X_t+\left(1-\frac {X_t} n\right)(X_t+1)
\\&=X_t+1-\frac {X_t} n
\\&=\left(1-\frac1n\right)X_t+1
\end{align}
Using the law of total expectation, the unconditional expectation of $X_t$ satisfies
\begin{align}
E (X_{t+1}) &= E(E (X_{t+1}|X_t))
\\&=E\left(\left(1-\frac1n\right)X_t+1\right)
\\&=\left(1-\frac1n\right)E X_t+1.
\end{align}
With the initial condition $X_0=0$, the solution of this first order linear non-homogenous difference equation is
$$
E X_t=\left[1 - \left(1-\frac1n\right)^t \right]n.
$$
Thus, for $n=10$ fishes, the expected number of fish marked after $t=7$ rounds would be
$$
E X_7=(1-0.9^7)10=5.217031.
$$
