Let's solve the given problem based on the probability distribution table you provided:

We need to compute the following probabilities:

a. $P(4 \leq X \leq 5)$

This asks for the probability that $X$ is between 4 and 5, inclusive.
$P(4 \leq X \leq 5) = P(X = 4) + P(X = 5)$ $P(4 \leq X \leq 5) = 0.17 + 0.13 = 0.30$

Answer: 0.30

b. $P(2 \leq X < 4)$

This asks for the probability that $X$ is between 2 and 4, where 2 is included, but 4 is not.
$P(2 \leq X < 4) = P(X = 2) + P(X = 3)$ $P(2 \leq X < 4) = 0.20 + 0.34 = 0.54$

Answer: 0.54

c. $P(X \leq 5)$

This asks for the cumulative probability that $X$ takes any value from 0 to 5, inclusive.
$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$ $P(X \leq 5) = 0 + 0.12 + 0.20 + 0.34 + 0.17 + 0.13 = 0.96$

Answer: 0.96

If anything is unclear or you'd like more detailed steps, feel free to ask!

5 Related Questions:

1. What is the total probability sum for any valid probability distribution?
2. How would the probabilities change if the die was fair (uniform distribution)?
3. How do you calculate $P(3 < X \leq 6)$ using this table?
4. Can we treat this die as biased, and how does that affect expected value calculations?
5. What is the expected value $E(X)$ for this distribution?

Tip: Always make sure the sum of all probabilities in a distribution equals 1 to confirm it’s valid.