SOLUTION: A business office orders paper supplies from one of three vendors, V1, V2, or V3. Orders are to be placed on two successive days, one order per day. Thus, (V2,V3) might denote th

Question 1184146: A business office orders paper supplies from one of three vendors, V1,
V2, or V3. Orders are to be placed on two successive days, one order
per day. Thus, (V2,V3) might denote that vendor V2 gets the order on
the first day and vendor V3 gets the order on the second day.
a. List the sample points in this experiment of ordering paper on two
successive days.
b. Assume the vendors are selected at random each day and assign a
probability to each sample point.
c. Let A denote the event that the same vendor gets both orders and B
the event that V2 gets at least one order. Find P(A), P(B), P(A∪ B)and
P(A∩ B)by summing the probabilities of the sample points in these
events
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
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

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