Answer
a) ${\bf 5\times 10^{-3}}$
b) ${\bf 2.5\times 10^{-3}}$
c) ${\bf 0}$
d) ${\bf 2.5\times 10^{-3}}$
Work Step by Step
We know that the probability of detecting an electron is given by
$$\text{Prob}( \text{in } \delta x \text{ at } x)=P(x)\delta x $$
$$\color{blue}{\bf [a]}$$
When $\delta x=0.010$ mm, and at $x=0$ mm, we can see from the given graph that $P(0)=0.50$ mm$^{-1}$
So,
$$\text{Prob}( \text{in 0.01-mm } \text{ at 0-mm})=(0.50)(0.01)=\color{red}{\bf 5\times 10^{-3}}$$
$$\color{blue}{\bf [b]}$$
When $\delta x=0.010$ mm, and at $x=0.5$ mm, we can see from the given graph that $P(0.5)=0.250$ mm$^{-1}$
So,
$$\text{Prob}( \text{in 0.01-mm } \text{ at 0.5-mm})=(0.250)(0.01)=\color{red}{\bf 2.5\times 10^{-3}}$$
$$\color{blue}{\bf [c]}$$
When $\delta x=0.010$ mm, and at $x=1$ mm, we can see from the given graph that $P(1)=0$ mm$^{-1}$
So,
$$\text{Prob}( \text{in 0.01-mm } \text{ at 1.0-mm})=(0)(0.01)=\color{red}{\bf 0}$$
$$\color{blue}{\bf [d]}$$
When $\delta x=0.010$ mm, and at $x=2$ mm, we can see from the given graph that $P(2)=0.25$ mm$^{-1}$
So,
$$\text{Prob}( \text{in 0.01-mm } \text{ at 2.0-mm})=(0.25)(0.01)=\color{red}{\bf2.5\times 10^{-3}}$$
