a. Since any real number in $[1,2)$ is a possible outcome, the sample space is indeed $S=[1,2)$.

b. Now, let's look at $P(1.5) .$ A reasonable guess would be $P(1.5)=0 .$ But can we provide a reason for that? Let us divide the $[1,2)$ interval to $2 N+1$ equal-length and disjoint intervals, $\left[1,1+\frac{1}{2 N+1}\right),\left[1+\frac{1}{2 N+1}, 1+\frac{2}{2 N+1}\right), \cdots,\left[1+\frac{N}{2 N+1}, 1+\frac{N+1}{2 N+1}\right), \cdots,\left[1+\frac{2 N}{2 N+1}, 2\right)$

The only information that we have is that the arrival time is "uniform" on the $[1,2)$ interval. Therefore, all of the above intervals should have the same probability, and since their union is

$S$ we conclude that

$$

\begin{array}{c}

P\left(\left[1,1+\frac{1}{2 N+1}\right)\right)=P\left(\left[1+\frac{1}{2 N+1}, 1+\frac{2}{2 N+1}\right)\right)=\cdots \\

\cdots=P\left(\left[1+\frac{N}{2 N+1}, 1+\frac{N+1}{2 N+1}\right)\right)=\cdots \\

\cdots=P\left(\left[1+\frac{2 N}{2 N+1}, 2\right)\right)=\frac{1}{2 N+1}

\end{array}

$$

In particular, by defining $A_{N}=\left[1+\frac{N}{2 N+1}, 1+\frac{N+1}{2 N+1}\right)$, we conclude that

$$

P\left(A_{N}\right)=P\left(\left[1+\frac{N}{2 N+1}, 1+\frac{N+1}{2 N+1}\right)\right)=\frac{1}{2 N+1}

$$

Now note that for any positive integer $N, 1.5 \in A_{N} .$ Thus, $\{1.5\} \subset A_{N}$, so

$$

P(1.5) \leq P\left(A_{N}\right)=\frac{1}{2 N+1}

$$

for all

Note that as $N$ becomes large, $P\left(A_{N}\right)$ approaches $0 .$ Since $P(1.5)$ cannot be negative, we conclude that $P(1.5)=0$. Similarly, we can argue that $P(x)=0$ for all $x \in[1,2)$.

c. Next, we find $P([1,1.5))$. This is the first half of the entire sample space $S=[1,2)$ and because of uniformity, its probability must be $0.5$. In other words,

$$

\begin{array}{ll}

P([1,1.5))=P([1.5,2)) & \text { (by uniformity) } \\

P([1,1.5))+P([1.5,2)) & =P(S)=1

\end{array}

$$

Thus

$$

P([1,1.5))=P([1.5,2))=\frac{1}{2}

$$

d. The same uniformity argument suggests that all intervals in $[1,2)$ with the same length must have the same probability. In particular, the probability of an interval is proportional to its length. For example, since

$$

[1,1.5)=[1,1.25) \cup[1.25,1.5)

$$

Thus, we conclude

$$

\begin{aligned}

P([1,1.5))=& P([1,1.25))+P([1.25,1.5)) \\

&=2 P([1,1.25))

\end{aligned}

$$

And finally, since $P([1,2))=1$, we conclude

$$

P([a, b])=b-a, \quad \text { for } 1 \leq a \leq b<2 .

$$