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Your friend tells you that she will stop by your house sometime after or equal to 1 p.m. and before 2 p.m., but she cannot give you any more information as her schedule is quite hectic. Your friend is very dependable, so you are sure that she will stop by your house, but other than that we have no information about the arrival time. Thus, we assume that the arrival time is completely random in the 1 p.m. and 2 p.m. interval. (As we will see, in the language of probability theory, we say that the arrival time is "uniformly" distributed on the $[1,2)$ interval). Let $T$ be the arrival time.

a. What is the sample space $S$ ?
b. What is the probability of $P(1.5) ?$ Why?
c. What is the probability of $T \in[1,1.5) ?$
d. For $1 \leq a \leq b \leq 2$, what is $P(a \leq T \leq b)=P([a, b])$ ?
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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
$$P([a, b])=b-a, \quad \text { for } 1 \leq a \leq b<2 .$$

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