The expectation of a number obtained when throwing a dice having 1 written on three faces, 2 on two faces and 5 on one face is given by

2 \(\frac{3}{2}\) \(\frac{71}{30}\) 6

2

$P(1)=\frac{1}{2}, P(2)=\frac{1}{3},P(5)=\frac{1}{6}$ $∴ E=1×\frac{1}{2}+2×\frac{1}{3}+5×\frac{1}{6}=\frac{3+4+5}{6}=\frac{12}{6}=2$ Option A is correct.