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.
The problem is that seeing $RR$ in the first two draws causes us to revise our estimates for the probabilities.
Assume that our prior was that the five states had equal probability (maybe not a great assumption, but I don't see how else to proceed). Then the total probability that you draw $RR$ initially is $$\frac 15\times \left(\sum_{i=2}^4\frac {\binom i2}{\binom 42}\right) = \frac 13$$
We use Bayes Theorem to revise our probability estimates and we get the new probabilities $$P(PPRR)=.1\quad P(PRRR)=.3\quad P(RRRR)=.6$$
Where do these numbers come from? Well, each of them just represents that portion of the total probability which is explained by being in the specified state. Thus, for example, $$P(PPRR)=\frac {\frac 15\times \frac {\binom 22}{\binom 42}}{\frac 13}=\frac 35 \times \frac 1{6}=\frac 1{10}$$
We then can read off the answer $$\frac {1}{10}\times \frac 12+\frac {3}{10}\times \frac 14+\frac 6{10}\times 0=\boxed {\frac 18}$$
Best Answer
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)
$$P = P_1 +P_2 = \frac{32}{100} = \frac{8}{25}$$