This is the question:
There are two resistors with resistance values $R_1=100\pm3$ ohm and $R_2=200\pm4$ ohm. Find the equivalent resistance of parallel combination.
According to what I've learnt, in any expression of multiplication or division, the percentage errors of each term are added up to find the equivalent percentage error. That is, if $$y=\frac{\text {AB}}{\text C}$$ then $$\%\;\text{error in y}=\%\;\text{error in A}+\%\;\text{error in B}+\%\;\text{error in C}$$
For the above problem, let $R_s$ denote series combination. Then $R_s=300\pm7$ ohm.
Let $R_p$ denote parallel combination.
$$\therefore R_p=\frac{R_1R_2}{R_1+R_2}=\frac{R_1R_2}{R_s}$$
Ignoring errors, we get $R_p=\frac{200}{3}$ ohm $=66.67$ ohm
$\%\;\text{error in R}_1=3$, $\%\;\text{error in R}_2=2$, $\%\;\text{error in R}_s=\frac73$
Hence, $\%\;\text{error in R}_p= 3+2+\frac73=\frac{22}{3}$
So, error in $R_p$ will be $\frac{22}{3}\%$ of $\frac{200}{3}$, which is approximately $4.89$.
Hence, I got $R_p=66.67\pm4.89$ ohm.
However, the book used the formula described and proved here and arrived at the answer $R_p=66.67\pm1.8$ ohm.
So, is the percentage error method wrong?