The discrete random variable T has probability function.
 𝑡 
0 
1 
2 
3

P(𝑇 = 𝑡) 
3𝑎 
2𝑎 
a
b

 
Find 𝑃(0.5 < 𝑇 ≤ 3).

To find \( P(0.5 < T \leq 3) \), we need to sum the probabilities of \( T \) taking the values that satisfy this condition. Here, the values that satisfy \( 0.5 < T \leq 3 \) are \( T = 1, 2, \text{and } 3 \).

Given the probability function:

\[
P(T = t) = 
\begin{cases} 
3a & \text{if } t = 0 \\
2a & \text{if } t = 1 \\
a & \text{if } t = 2 \\
b & \text{if } t = 3 
\end{cases}
\]

We need to find \( P(1) + P(2) + P(3) \).

\[
P(0.5 < T \leq 3) = P(T = 1) + P(T = 2) + P(T = 3)
\]

Substitute the given probabilities:

\[
P(0.5 < T \leq 3) = 2a + a + b = 3a + b
\]

Therefore, \( P(0.5 < T \leq 3) = 3a + b \).

Would you like more details or have any questions about this solution?

Here are 5 numbered questions you might ask next:
1. How do you find the values of \(a\) and \(b\) if the probabilities must sum to 1?
2. Can you explain the properties of a probability mass function (pmf)?
3. How do you calculate the expected value of a discrete random variable?
4. What is the variance of a discrete random variable and how is it computed?
5. Can you solve a problem involving the cumulative distribution function (CDF) of a discrete random variable?

**Tip:** When dealing with probability distributions, always ensure that the total probability sums to 1.

Learn how to calculate the probability P(0.5 < T ≤ 3) for a discrete random variable T with a given probability mass function. This problem involves understanding probability theory and the application of probability mass functions to find specific probabilities.

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