Question 1184146
Here's the solution:

**a. Sample Points:**

The sample points represent all possible pairs of vendors chosen on two successive days.  We can list them as ordered pairs (Vendor Day 1, Vendor Day 2):

* (V1, V1)
* (V1, V2)
* (V1, V3)
* (V2, V1)
* (V2, V2)
* (V2, V3)
* (V3, V1)
* (V3, V2)
* (V3, V3)

**b. Probabilities:**

Since the vendors are selected at random each day, each vendor has an equal probability of being chosen (1/3).  Since the orders on the two days are independent events, we can find the probability of each sample point by multiplying the probabilities for each day.

* P(V1, V1) = (1/3) * (1/3) = 1/9
* P(V1, V2) = (1/3) * (1/3) = 1/9
* P(V1, V3) = (1/3) * (1/3) = 1/9
* P(V2, V1) = (1/3) * (1/3) = 1/9
* P(V2, V2) = (1/3) * (1/3) = 1/9
* P(V2, V3) = (1/3) * (1/3) = 1/9
* P(V3, V1) = (1/3) * (1/3) = 1/9
* P(V3, V2) = (1/3) * (1/3) = 1/9
* P(V3, V3) = (1/3) * (1/3) = 1/9

**c. Probabilities of Events A and B:**

* **Event A (Same vendor gets both orders):**
   A = {(V1, V1), (V2, V2), (V3, V3)}
   P(A) = P(V1, V1) + P(V2, V2) + P(V3, V3) = (1/9) + (1/9) + (1/9) = 3/9 = 1/3

* **Event B (V2 gets at least one order):**
   B = {(V1, V2), (V2, V1), (V2, V2), (V2, V3), (V3, V2)}
   P(B) = P(V1, V2) + P(V2, V1) + P(V2, V2) + P(V2, V3) + P(V3, V2) = (1/9) + (1/9) + (1/9) + (1/9) + (1/9) = 5/9

* **Event A ∪ B (A or B occurs):**
   A ∪ B = {(V1, V1), (V2, V2), (V3, V3), (V1, V2), (V2, V1), (V2, V3), (V3, V2)}
   P(A ∪ B) = P(V1, V1) + P(V2, V2) + P(V3, V3) + P(V1, V2) + P(V2, V1) + P(V2, V3) + P(V3, V2) = (1/9) + (1/9) + (1/9) + (1/9) + (1/9) + (1/9) + (1/9) = 7/9
   *Alternatively*, since A is a subset of B, P(A∪B) = P(B) = 5/9. I made a mistake in the previous calculation.

* **Event A ∩ B (A and B occur):**
   A ∩ B = {(V2, V2)}
   P(A ∩ B) = P(V2, V2) = 1/9