Suppose that a probability of 116 is assigned to eachpoint of the sample space of part (a) of Exercise 3.1on page 65. Find the probability distribution of thetotal number of units of black and white cement thatare adulterated.
The Correct Answer And Explanation is:
To answer your question, I’ll guide you through solving it based on the context. The exercise you’re referring to comes from a probability textbook. Although I don’t have the exact book, I am familiar with common versions of such exercises in introductory probability texts like Sheldon Ross’s A First Course in Probability.
Assumptions Based on Typical Exercise 3.1:
Let’s assume Exercise 3.1 involves a sample space consisting of combinations of black and white cement units, each of which may or may not be adulterated. A typical setup is:
- There are 2 black and 2 white cement units.
- Each unit is either adulterated (A) or not adulterated (N).
- The sample space lists all combinations of (A/N) across the 4 units.
There are 24=162^4 = 16 total outcomes (each unit can independently be A or N), and the problem states that each outcome has a probability of 116\frac{1}{16}.
Random Variable:
Let XX be the total number of adulterated cement units (black and white).
Then X∈{0,1,2,3,4}X \in \{0, 1, 2, 3, 4\}, and we must determine the probability distribution of XX.
We count how many outcomes have exactly 0, 1, 2, 3, or 4 adulterated units.
- P(X=0)P(X = 0): All units are not adulterated → only 1 outcome.
- P(X=1)P(X = 1): Exactly one unit is adulterated → choose 1 out of 4 → (41)=4\binom{4}{1} = 4.
- P(X=2)P(X = 2): Choose 2 out of 4 to be adulterated → (42)=6\binom{4}{2} = 6.
- P(X=3)P(X = 3): Choose 3 out of 4 to be adulterated → (43)=4\binom{4}{3} = 4.
- P(X=4)P(X = 4): All 4 are adulterated → only 1 outcome.
Final Probability Distribution of XX:
| XX | Probability P(X)P(X) |
|---|---|
| 0 | 116\frac{1}{16} |
| 1 | 416\frac{4}{16} |
| 2 | 616\frac{6}{16} |
| 3 | 416\frac{4}{16} |
| 4 | 116\frac{1}{16} |
Explanation (300+ words):
In probability theory, a sample space consists of all possible outcomes of an experiment. In this case, we are analyzing the total number of adulterated units (both black and white cement) in a setup involving four cement units.
Each cement unit can independently be adulterated or not, so there are 2 choices (A or N) per unit. Since there are 4 units, the total number of combinations is 24=162^4 = 16. Each combination is assumed to be equally likely, meaning every outcome in the sample space has a probability of 116\frac{1}{16}.
We define a random variable XX as the total number of adulterated units. This is a discrete random variable because it can only take specific integer values: 0, 1, 2, 3, or 4.
To find the probability distribution of XX, we count how many of the 16 total outcomes correspond to each value of XX. For example, there is only 1 way to have 0 adulterated units (all 4 are not adulterated), and 1 way to have 4 adulterated units. There are 4 ways to have exactly 1 adulterated unit (any one of the four could be adulterated), 6 ways to have exactly 2 adulterated units (choose any 2 of 4), and again 4 ways to have exactly 3 adulterated units.
Once the counts are found, the probability for each value of XX is simply the count divided by 16. This gives us a full probability distribution, which is fundamental in understanding the likelihood of various numbers of adulterated units appearing in the sample.
This distribution is symmetrical, peaking at X=2X = 2, and represents a binomial distribution with n=4n = 4, p=0.5p = 0.5, due to the independent and equally likely adulteration.