The answer to the first question is correct.
What is the initial probability p(nS) that the bag contains nS silver coins?
$$P(nS) = 1/11 $$
Now, the second question
You pull out one coin: it is silver. What is now the probability that the bag contains 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 silver coins?
The analysis in the question is wrong because the question is about conditional probability. We want to find out the probability of the event that the bag contains n silver coins($0 \le n \le 9$), given that the first coin pulled out was a silver coin. We will use Bayes' theorem to solve this.
See this and this for Bayes theorem and this for an illustrative example. The idea of drawing all the probabilities as a tree demonstrated in this idea is quite helpful in understanding the problem. Now, solution to the problem.
Lets us call the event we pulled out one silver coin as $O$.
Let us call the event there are n silver coins remaining in the bag as $nS$, where $0 \le n \le 9$
We want to find out $P(nS | O)$
Using Bayes formula
$$P(nS|O) = \frac{P(O|nS). P(nS)}{P(O)}$$
From the above question, we know that
$$P(nS) = 1/11 $$
$P(O|nS)$ is defined as the probability where a silver coin is picked given there are n silver coins remaining in the bag after the ball is picked. Since there are n silver coins remaining after the first silver coin was picked, the probability $P(O|nS)$ can be defined as $\frac{n+1}{10}$
$$P(O|nS) = \frac{n+1}{10}$$
$P(O)$ is the probability that a silver coin is picked from the bag on the first turn. Since there are either silver or gold coins in the bag, there is equal probability for both. Thus,
$$P(O) = 1/2$$
This can also be derived by drawing the tree as illustrated in the video
$$P(O) = (\frac{1}{11}*\frac{0}{10}) + (\frac{1}{11}*\frac{1}{10}) + .... + (\frac{1}{11}*\frac{9}{10}) + (\frac{1}{11}*\frac{10}{10})$$
$$P(O) = \frac{1}{11}(\frac{0}{10} + \frac{1}{10} + .... + \frac{9}{10} + \frac{10}{10})$$
$$P(O) = \frac{1}{11}(\frac{0+1+2+ ... + 9 + 10}{10}) = \frac{55}{11*10} = \frac{1}{2}$$
Substituting these values in
$$P(nS|O) = \frac{P(O|nS). P(nS)}{P(O)}$$
$$P(nS|O) = \frac{n+1}{55}$$
You now pull out another coin: it is gold. What is now the probability that the bag contains 0, 1, 2, 3, 4, 5, 6, 7, 8 silver coins?
Let us call the event where two coins were pulled, and the first one is silver and other is gold as $T$.
Let us call the event there are n silver coins remaining in the bag as $nS$, where $0 \le n \le 8$
We want to find out $P(nS | T)$
Using Bayes formula
$$P(nS|T) = \frac{P(T|nS). P(nS)}{P(T)}$$
We know that
$$P(nS) = 1/11 $$
$P(T|nS)$ is defined as the probability where a silver coin is drawn first and gold coin is drawn next given there are n silver coins remaining in the bag after the two coins are picked. On the first turn, the probability for a silver coin to be picked is $\frac{n+1}{10}$. In the second turn, 9 coins would be left in the bag and n of them are silver coins, so, there would be $9-n$ gold coins. The probability of picking a gold coin in the second turn would be $\frac{9-n}{9}$. Combined, the probability would be
$$P(T|nS) = \frac{n+1}{10}*\frac{9-n}{9}$$
$P(T)$ is the probability that a black ball is picked from the bag first followed by a white ball. This can be derived from the probability tree
$$P(T) = (\frac{1}{11}*\frac{0}{10}) + (\frac{1}{11}*\frac{1}{10}*\frac{9}{9}) + (\frac{1}{11}*\frac{2}{10}*\frac{8}{9}) .... + (\frac{1}{11}*\frac{9}{10}*\frac{1}{9}) + (\frac{1}{11}*\frac{10}{10}*\frac{0}{9})$$
$$P(T) = \frac{1}{11}(\sum_{n=0}^{10} \frac{n(10-n)}{10*9})$$
$$P(T) = \frac{1}{11}(\frac{165}{90}) = \frac{1}{6}$$
Substituting these values in
$$P(nS|T) = \frac{P(T|nS). P(nS)}{P(T)}$$
$$P(nS|T) = \frac{8n+9-n^2}{165}$$
Let the probability that coin drawn from the first bag is $1,2,3,4$ be $p_1$. We can quickly see that
$$p_1 = \frac{4}{10}$$
Now the probability that coin drawn from the second bag is $7,8,9,10$ be $p_2$. We can see again that this probability is
$$p_2 = \frac{4}{10}$$
So net probability will be a product of $p_1$ and $p_2$ as they are independent events and both need to happen simultaneously.
$$P_1 = \frac{4}{10}\cdot \frac{4}{10} \frac{16}{100}$$
Now we take the second case that the coin drawn from first bag is $1,2,3,4$ and coin from the second bag is $7,8,9,20$. Since both bags are identical, this gives us the probability same as before
$$P_2 = \frac{4}{10}\cdot \frac{4}{10} \frac{16}{100}$$
Summing up $P_1$ and $P_2$ (as we need to find the union and they are mutually exclusive)
$$P = P_1 +P_2 = \frac{32}{100} = \frac{8}{25}$$
Best Answer
To understand the formula, it would be easiest to explain how it works conceptually before we derive it.
Let's simplify the problem and say there are only 3 bags each with 2 coins in them. 2 of those bags have the 1 gram coins and one has the 1.01 gram gold coins. Let's denote the bags arbitrarily as $Bag_0$, $Bag_1$, and $Bag_2$. Similarly to your problem, let's take 0 coins from $Bag_0$, 1 coin from $Bag_1$, and 2 coins from $Bag_2$. We know that the gold coins must be in one of those bags, so there are three possibilities when we weigh the three coins we removed:
Gold Coins in $Bag_0$: So the weight of the 3 coins on the scale are all 1 gram. So the scale will read 3 grams.
Gold Coins in $Bag_1$: So the weight of 1 of the coins is 1.01 grams and 2 of the coins are 2 grams. So the scale will read 3.01 grams.
Gold Coins in $Bag_2$: So the weight of 2 of the coins is 2.02 grams and 1 of the coins is 1 gram. So the scale will read 3.02 grams.
So each possibility has a unique scenario. So if we determine the weight, we can determine from which bag those coins came from based on that weight.
We can generalize our results from this simplified example to your 100 bag example.
Now for deriving the formula. Say hypothetically, of our 100 bags, all 100 coins in each of the 100 bags weigh 1 gram each. In that case, when we remove 0 coins from $Bag_0$, 1 from $Bag_1$, up until 99 coins from $Bag_{99}$, we'll have a total of 4950 coins on the scale, which will equivalently be 4950 grams. Simply put, if $n$ is our Bag number (denoted $Bag_n$), we've placed $n$ coins from each $Bag_n$ onto the scale for $n = 0, 1, 2, ... 99 $.
So the weight of the coins will be $Weight = 1 + 2 + 3 + ... + 99 = 4950$
But we actually have one bag with gold coins weighing 1.01 grams. And we know that those 1.01 gram coins must be from some $Bag_n$. In our hypothetical example, all of our coins were 1 gram coins, so we must replace the $n$ coins weighed from $Bag_n$ with $n$ gold coins weighing 1.01 grams. Mathematically, we would have: $Weight = 4950 - n + 1.01n = 4950 + .01n = 4950 + n/100$
Rearranging the formula to solve for n, we have: $100(Weight-4950) = n$, where $Weight$ is $W$ and $n$ is $N$ in your example.
I have no knowledge of an alternative answer to this puzzle, but perhaps another member's answer may be enlightening if there is. Technically speaking, you could have denoted the bags from 1 to 100 and gone through a similar process as above, but the method is still the same, so I wouldn't treat it as a new answer.
If our electric scale is replaced by a scale of libra, I don't believe it would be possible to answer this puzzle with only one measurement of weight. But again, perhaps another answer may be enlightening on that.