Correct answer is 71
\(\mathrm{a}=1-\frac{{ }^{3} \mathrm{C}_{5}}{{ }^{12} \mathrm{C}_{5}}\)
\(\mathrm{b}=3 \cdot \frac{{ }^{9} \mathrm{C}_{4}}{{ }^{12} \mathrm{C}_{5}}\)
\(\mathrm{c}=3 \cdot \frac{{ }^{9} \mathrm{C}_{3}}{{ }^{12} \mathrm{C}_{5}}\)
\(\mathrm{d}=1 \cdot \frac{{ }^{9} \mathrm{C}_{2}}{{ }^{12} \mathrm{C}_{5}}\)
\(\mathrm{u}=0 . \mathrm{a}+1 . \mathrm{b}+2 . \mathrm{c}+3 . \mathrm{d}=1.25\)
\(\sigma^{2}=0 . a+1 . b+4 . c+9 d-u^{2}\)
\(\sigma^{2}=\frac{105}{176}\)
\(176-105=71\)