AP ICET 2017 Shift 1 Paper

Total number of outcomes $=100$
Number of numbers divisible by 3 $=33$
Number of numbers divisible by 5 $=20$
Number of numbers divisible by $\left(3\times 5\right)$ i.e., $=6$
∴ Number of favourable outcomes$=33+20\mathit{–}6=47$
∴ Required probability $=\frac{47}{100}$