Finding Various Probabilities based on given Cumulative Density Function

Question

Given the CDF below, I am looking to calculate the probability of the following four events:

1. $\{X = 2\}$
2. $\{X < 2\}$
3. $\{X = 2\} \cup \{0.5 \leq X \leq 1.5\}$
4. $\{X = 2\} \cup \{0.5 \leq X \leq 3\}$

The random variable X for the given CDF is neither continuous or discrete, and that is why I am having issues calculating the aforementioned probabilities.

I believe I understand the first problem. Based on the CDF, the random variable $X$ takes on values $X \in \{[0, 1] \cup [2, 4]\}$. To calculate $P(X = 2)$, We can do $P(X \leq 2) - P(X \leq 1) = F(2) - F(1) = 1/3$.

I believe I also understand the second problem, as $P(X < 2)$ simply is $P(X \leq 1) = F(1) = 1/3$

The final two problems are where I am particularly struggling, as I am unsure of how to calculate $P(0.5 \leq X \leq 1.5)$ and $P(0.5 \leq X \leq 3)$.

If anyone could confirm and/or correct my thinking for problems $1$ and $2$ as well as provide some insight for the final $2$ problems, it would be much appreciated.

Answer 1

The first two parts are good.

Using similar principle, $$ P(0.5 \leq X \leq 1.5) = P(X \leq 1.5) - P(X < 0.5) $$ Note that $$ F(x) = P(X \leq x) = P(X < x) + P(X = x)$$

Therefore when $F$ is continuous at $x$, i.e. the graph $y = F(x)$ has no discrete jump like the one at $x = 2$, then we know that $P(X = x)$ vanish and thus in such case $$ P(X < x) = F(x) $$

Put into above result $$ P(0.5 \leq X \leq 1.5) = F(1.5) - F(0.5)$$

The other question is similar. Note that the graph $y = F(x)$ is a straight line from $x = 2$ to $x = 4$, and $3$ is the midpoint of $[2, 4]$, therefore $F(3)$ is the midpoint of $[F(2), F(4)]$

Hope it helps.

Edit: Also note that for disjoint events $A, B$,

$$ P(A \cup B) = P(A) + P(B) $$

Instead, if $A \subseteq B$, then $A \cup B = B$ and thus in this case

$$ P(A \cup B) = P(B) $$

Answer 2

Looking at the graph $P(X=2)=\frac{1}{3}$ and $P(X\lt2)=\frac{1}{3}$, so your approach is correct.
For the third problem you need $P(X\le 1.5)-P(\lt .5)=\frac{1}{3}-\frac{1}{12}=\frac{1}{4}$ resulting as $\frac{7}{12}$, since the pieces are disjoint.
For (4) you need to be careful since $x=2$ is inside $.5\le X\le 3$, contributing nothing to the union. That probability is $P(X\le 3)-P(X\lt .5)=\frac{5}{6}-\frac{1}{12}=\frac{3}{4}$