Rounding Error

Let the initial number be $n$ and it after rounded to two decimal places be $m$ . Then we need $m \in [0.15, 0.25)$ in order to be rounded to 0.2, which means $m \in [0.15, 0.24]$ as $m$ has two decimal places. This means $n \in [0.145, 0.245)$ ; the value $0.145$ is rounded to $0.15$ , and the value near $0.245$ is rounded to $0.24$ . But since we pick $n$ from the interval $[0.15, 0.25]$ , only their intersection matters: we round to 0.2 if and only if $n \in [0.15, 0.245)$ .

Assuming $n$ is uniformly distributed on this interval, the chance of it lying in $[0.15, 0.245)$ is equal to its length, $0.095$ , divided by the length of the interval where we sample $n$ from $[0.15, 0.25])$ , namely $0.1$ . This is $\boxed{95}\%$ .

Let $x\in[0.15;\:0.25]=:\Omega\subset\mathbb{R}$ be the randomly chosen number and $x_1,\:x_2$ the results after the first and second rounding process, respectively. Note that after the first rounding, $x_1\in \Omega$ . We are interested in $\begin{aligned} P(x_2=0.2) &= P(x_1\in[0.15;\:0.25\red{)}) \underset{x_1\in \Omega}{=} 1 - P(x_1=\red{0.25}) \\[1em] &= 1 - P(x\in[0.245;\:0.25]) = 1 - \frac{0.25-0.245}{0.25-0.15}=0.95=\boxed{95}\% \end{aligned}$

In the last step, we can calculate the probability by comparing the lengths of $\Omega$ and $[0.245;\:0.25]$ .