Question 1178055: An experiment consists of drawing four objects from a container, which holds eight operable, six defective, and 10 semioperable objects. Let X be the number of operable objects drawn and Y the number of defective objects drawn.
(a) Find the joint probability function of the bivariate random variable (X, Y).
(b) Find P(X = 3, Y = 0).
(c) Find P(X < 3, Y = 1).
(d) Give a graphical presentation of (a), (b), and (c).
Thank you :)
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step-by-step.
**Understanding the Problem**
* **Total Objects:** 8 (operable) + 6 (defective) + 10 (semioperable) = 24
* **Objects Drawn:** 4
* **X:** Number of operable objects drawn
* **Y:** Number of defective objects drawn
**a. Joint Probability Function P(X = x, Y = y)**
We need to find the probability of drawing x operable objects and y defective objects, given that we draw 4 objects in total.
* **Number of Semioperable Objects Drawn:** 4 - x - y
* **Total Combinations:** C(24, 4) = 24! / (4! * 20!) = 10626
* **Combinations of Operable Objects:** C(8, x)
* **Combinations of Defective Objects:** C(6, y)
* **Combinations of Semioperable Objects:** C(10, 4 - x - y)
The joint probability function is:
* P(X = x, Y = y) = [C(8, x) * C(6, y) * C(10, 4 - x - y)] / C(24, 4)
Where:
* x can be 0, 1, 2, 3, 4
* y can be 0, 1, 2, 3, 4
* 0 ≤ x + y ≤ 4
**b. P(X = 3, Y = 0)**
* P(X = 3, Y = 0) = [C(8, 3) * C(6, 0) * C(10, 1)] / C(24, 4)
* C(8, 3) = 8! / (3! * 5!) = 56
* C(6, 0) = 1
* C(10, 1) = 10
* P(X = 3, Y = 0) = (56 * 1 * 10) / 10626 = 560 / 10626 ≈ 0.0527
**c. P(X < 3, Y = 1)**
We need to calculate P(X = 0, Y = 1) + P(X = 1, Y = 1) + P(X = 2, Y = 1).
* **P(X = 0, Y = 1):** [C(8, 0) * C(6, 1) * C(10, 3)] / C(24, 4) = (1 * 6 * 120) / 10626 = 720 / 10626 ≈ 0.0678
* **P(X = 1, Y = 1):** [C(8, 1) * C(6, 1) * C(10, 2)] / C(24, 4) = (8 * 6 * 45) / 10626 = 2160 / 10626 ≈ 0.2033
* **P(X = 2, Y = 1):** [C(8, 2) * C(6, 1) * C(10, 1)] / C(24, 4) = (28 * 6 * 10) / 10626 = 1680 / 10626 ≈ 0.1581
* **P(X < 3, Y = 1) = 0.0678 + 0.2033 + 0.1581 = 0.4292**
**d. Graphical Presentation**
**(a) Joint Probability Function Table**
We can create a table showing the probabilities for various values of X and Y.
| X\Y | 0 | 1 | 2 | 3 | 4 |
|-----|----------|----------|----------|----------|----------|
| 0 | 0.0099 | 0.0678 | 0.0283 | 0.0022 | 0 |
| 1 | 0.0475 | 0.2033 | 0.0847 | 0.0071 | 0 |
| 2 | 0.1188 | 0.1581 | 0.0494 | 0.0035 | 0 |
| 3 | 0.0527 | 0.0413 | 0.0098 | 0.0003 | 0 |
| 4 | 0.0028 | 0.0011 | 0.0001 | 0 | 0 |
**(b) P(X = 3, Y = 0)**
This is a single point on the graph. It's a bar at (3, 0) with height 0.0527.
**(c) P(X < 3, Y = 1)**
This involves summing the probabilities for (0, 1), (1, 1), and (2, 1). We can represent this graphically as a sum of bars at these points.
**Graphical Representation Notes:**
* For the joint probability function, you'd typically use a 3D bar chart or a heatmap to visualize the probabilities for different combinations of X and Y.
* For (b) and (c), you could use a bar chart or simply mark the points on a graph.
* Due to the complexity of the data, a software like python matplotlib or R is recommended for the graphical presentation.
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