Calculating uncertainty when combining parallel resistors

The largest value of $R$ will be when $R_1$ and $R_2$ have their largest values. So putting $R_1= 132\ \Omega$ and $R_2= 550\ \Omega$, we get 106.5 $\Omega$, whereas if we put in the 'central' values, $R_1= 120\ \Omega$ and $R_2= 500\ \Omega$, we get 97 $\Omega$. The discrepancy as a percentage of 97 $\Omega$, is 10%.

The moral: elementary methods are sometimes the best! You can do this calculation by compounding individual uncertainties:

The percentage uncertainties in $\frac{1}{R_1}$ and $\frac{1}{R_2}$ are the same as in $R_1$ and $R_1$ because taking the reciprocal is a division. So, as Michael Seifert has calculated,$$G_1=\frac{1}{R_1}=(8.33±0.83)\times 10^{-3} \Omega^{-1}$$ $$G_2=\frac{1}{R_2}=(2.00±0.20)\times 10^{-3} \Omega^{-1}$$

We now have to add these conductances, so (in this naïve treatment) we add the absolute uncertainties. [We'd also add absolute uncertainties if we were subtracting quantities.] $$\text{So}\ \ \ \ G=G_1+G_2=(10.33±1.03)\times 10^{-3} \Omega^{-1}$$

The percentage uncertainty in $G$ is therefore 10%. Taking the final reciprocal (a division operation) leaves the percentage uncertainty as 10%. $$\text{So}\ \ \ \ R=\frac{1}{G}=(97±10) \Omega.$$

For divisions and multiplications you add percentage uncertainties, so if there is only one variable, as in $\frac{1}{R_1},$ the percentage uncertainty stays as it is. For quantities that are added or subtracted you add absolute uncertainties. For multiplication and division the addition of percentage uncertainties is, strictly, a first order approximation, but the whole treatment is oversimplified anyway (see M Seifert's answer).

There is no mathematical reason to give a 10% percent error on the final resistance; there is, however, and engineering one.

First: If the percentage number encoded in the resistors' stripes were a 1-sigma Gaussian error, then @Michael Seifert's answer would be correct: add the resistors harmonically and the errors are reduced in quadrature.

But: Those number are not errors, they are tolerances. So $R=100\pm10\,\Omega$ is supposed to mean:

$$ R({\rm Ohm}) \in [90, 110] $$

In which case, @Philip Wood's answer is a better answer: to get the range, use the range.

But there is a problem: there are 5% and 1% resistors too. And they aren't included in the 10% stock. Hence, your 100 Ohm resistors exclude any ones that satisfies:

$$ R({\rm Ohm}) \in [95, 105] $$

which makes for bimodal input distributions (this is simply a terrible exam question, with a wrong answer, btw).

At this point, you have to Monte Carlo it, with the result that:

$$ R = 96.6\pm 6.2\,\Omega $$

and

$$ 87 < R < 106.5 $$

See figure.

(note: a probability density per Ohm is in fact measured in Siemens ;-)