Calculating transmittance LED
You sure you use the correct equations? Because for normal incidence, it is:
$$R={\left\lvert\frac{n_1-n_2}{n_1+n_2}\right\rvert\\}^2$$
$$T=1-R$$
Reference - Fresnel Equation for Reflectance
Based on above equations, for GaAs-Glass interface, you will get -
$$R_1=0.15$$
$$T_1=0.85$$
For Glass-Air interface, you will get -
$$R_2=0.04$$
$$T_2=0.96$$
So mathematically, the effective transmittance should be -
$$T_{eff}=T_1 \times T_2 = 0.816 = 81.6 \% $$
therefore effective reflectance should be -
$$R_{eff}=1-T_{eff}=0.184=18.4\%$$
If you plug in the relationship between \$R\$ and \$T\$, you can derive relation for effective reflectance as well -
$$R_{eff}=1-T_{eff}$$
$$R_{eff}=1-(T_1\times T_2)$$
$$R_{eff}=1-((1-R_1)\times (1-R_2))$$
$$\implies R_{eff}=R_1+R_2-R1.R2$$