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:
Thanks for the question! However, to give you an accurate answer and detailed explanation, I need a bit more context about Exercise 3.1 on page 65 and what part (a) involves. Without that, I can’t be sure about the sample space, nor the specifics of the probability assignment or the structure of the problem.
Could you please provide the text or details of Exercise 3.1 part (a) on page 65? For example:
- What is the sample space?
- What is the context (cement, black and white units, adulteration counts)?
- What does “probability of 116 assigned to each point” mean exactly? (Is it 1/16, 1/6, or something else?)
General Approach to Such a Problem (Assuming Typical Setup)
Suppose we have two categories: black cement and white cement units. Each unit may or may not be adulterated. The total number of adulterated units can range from 0 up to the total number of units sampled.
If the sample space consists of all possible outcomes (combinations of adulterated black and white units), and each outcome is assigned an equal probability, then:
- The probability of each outcome = 1 / (total number of outcomes).
- To find the probability distribution of the total number of adulterated units = sum over all outcomes where total adulterated units = k.
Step-by-step method to find the distribution:
- Identify the sample space and outcomes:
For example, if you have 2 black cement units and 2 white cement units, each can be either adulterated (A) or not (N). Then sample points look like: (A, N, A, N), etc. - Calculate total number of outcomes:
For 4 units, each with 2 states → 2^4 = 16 outcomes. - Assign equal probability to each outcome:
Each outcome probability = 1/16. - Count outcomes for each total adulterated units k (from 0 to 4):
For example:- k=0 adulterated: only 1 outcome (N, N, N, N)
- k=1 adulterated: 4 outcomes
- k=2 adulterated: 6 outcomes
- k=3 adulterated: 4 outcomes
- k=4 adulterated: 1 outcome
- Calculate probabilities for each k:
Probability(total adulterated = k) = (number of outcomes with k adulterated) × (1/16).
Example with the above assumptions:
| Total adulterated units (k) | Number of outcomes | Probability P(X=k) = outcomes/16 |
|---|---|---|
| 0 | 1 | 1/16 = 0.0625 |
| 1 | 4 | 4/16 = 0.25 |
| 2 | 6 | 6/16 = 0.375 |
| 3 | 4 | 4/16 = 0.25 |
| 4 | 1 | 1/16 = 0.0625 |
This distribution is basically the binomial distribution with parameters n=4 and p=0.5 (assuming equal chance for adulteration or not).
Explanation
In probability problems involving a finite number of outcomes with equal probability, each individual outcome is assigned a probability equal to 1 divided by the total number of outcomes in the sample space. When the problem involves counting how many units are adulterated in two categories (black and white cement), each possible outcome (e.g., black unit 1 adulterated or not, white unit 1 adulterated or not) can be viewed as a binary event.
By calculating the total number of possible outcomes (2^n where n is the total units), we distribute the total probability evenly among these outcomes. Then, by grouping outcomes according to the total number of adulterated units (adding the adulterated counts from black and white), we can compute the probability that exactly k units are adulterated.
This gives the probability distribution of the random variable “total number of adulterated units”. The distribution in this simplified example follows a binomial pattern, reflecting the combination of independent binary events.