A box contains $a$ batteries of which $d$ are dead. The batteries are tested randomly, one by one. Every time that a good battery is drawn, it is returned to the box. When a dead battery is drawn, it is replaced by a good one. Let $X_n$ denote the number of good batteries in the box after $n$ of them are checked.
(i) Find an expression for the random variable $E(X_n\mid X_{n-1})$ and show that: $E(X_n) = 1 + (1-\frac 1a)E(X_{n-1}), n≥1.$
(ii) Show by induction that: $E(X_n )=a-d(1-\frac 1a)^n$
Can someone please help me with this? This question (link under) is very similar but I still can't figure it out. expected value of battery drawing problem
For (i) note that $(X_n)_{n\in\mathbb N}$ is a Markov chain with state space $S := \{a-d, \ldots , a\}$ and transition probabilities $$p(x,y) = \begin{cases} \frac x a &: x = y \\ \frac {a-x} a &: x = y-1 \\ 0 & : \text{else}. \end{cases}$$ Thus we can write $$\mathbb E [X_n | X_{n-1}] = \sum_{y\in S} p(X_{n-1} , y) \ y = \frac {X_{n-1}^2} a + \frac{(a-X_{n-1})(X_{n-1}+1)}{a} \\ = 1 + \left( 1-\frac 1 a \right) X_{n-1}.$$ In conclusion $$\mathbb E [X_n] = \mathbb E [\mathbb E[X_n | X_{n-1}]] = \mathbb 1+ \left(1 -\frac 1 a \right) \mathbb E [X_{n-1}].$$ Since $X_0 = a - d$ statement (ii) follows easily by induction by using (i).