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 give you the correct answer and a thorough explanation, we must first reconstruct the context of Exercise 3.1(a) on page 65, which typically appears in probability textbooks such as “A First Course in Probability” by Sheldon Ross. Since you referenced black and white cement and sample space, let’s reconstruct the likely scenario based on common setups:
Assumed Setup of Exercise 3.1(a):
You randomly select two bags of black cement and two bags of white cement, and you test each one to determine whether it is adulterated or not. So the sample space consists of all possible outcomes in which each of the 4 bags (2 black + 2 white) can either be adulterated (A) or non-adulterated (N).
Since each of the 4 bags can be in 2 states, the total number of outcomes is: 24=162^4 = 16
Each outcome is equally likely with a probability of: P(each outcome)=116P(\text{each outcome}) = \frac{1}{16}
We define the random variable X = total number of adulterated bags (both black and white combined). Possible values for X are: X=0,1,2,3,4X = 0, 1, 2, 3, 4
We now count how many outcomes correspond to each value of X, and compute the corresponding probabilities.
Probability Distribution of X (Total Adulterated Bags)
| X | Number of Outcomes | P(X) |
|---|---|---|
| 0 | 1 | 1/16 |
| 1 | 4 | 4/16 = 1/4 |
| 2 | 6 | 6/16 = 3/8 |
| 3 | 4 | 4/16 = 1/4 |
| 4 | 1 | 1/16 |
Explanation
In this problem, we analyze the probability distribution of a random variable X, which represents the total number of adulterated cement bags in a sample consisting of two black and two white cement bags. Each bag can independently be either adulterated (A) or not adulterated (N). Since there are 4 bags and each has 2 possible states, the total number of possible combinations in the sample space is: 24=162^4 = 16
Each of these 16 outcomes is equally likely, with probability 116\frac{1}{16}. Our goal is to calculate the distribution of X, i.e., how many outcomes yield 0, 1, 2, 3, or 4 adulterated bags.
To find this, we systematically count the number of outcomes that result in each possible value of X:
- X = 0: All bags are non-adulterated. Only 1 outcome: NNNN.
- X = 1: Exactly one bag is adulterated. We can choose any 1 of the 4 positions to be A; the rest are N. There are 4 such outcomes.
- X = 2: Two adulterated bags. We choose any 2 of the 4 positions to be A. This is (42)=6\binom{4}{2} = 6 outcomes.
- X = 3: Three adulterated bags. Choose 3 out of 4 positions: (43)=4\binom{4}{3} = 4.
- X = 4: All four bags are adulterated. Only 1 outcome: AAAA.
Thus, the probabilities for each value of X are computed by dividing the number of favorable outcomes by 16. The result is a symmetric distribution centered at X = 2, which is the most probable value. This reflects the binomial-like nature of the problem with independent binary outcomes.
This analysis helps us understand how likely different levels of adulteration are in random samples, and is a fundamental technique in probability modeling and quality control.
