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 solve this, we need to reconstruct Exercise 3.1 (a) from the textbook context you’re referring to. Although I don’t have direct access to that textbook, the exercise is a classic probability problem involving cement bags (often involving black and white cement bags and identifying whether they are adulterated or not).
Assumed Context from Exercise 3.1(a):
Suppose:
- A company tests two bags, one of black and one of white cement.
- Each bag may either be adulterated (A) or not adulterated (N).
- The sample space consists of all possible combinations of adulteration status for the two bags.
Then, the sample space SS is: S={(A,A),(A,N),(N,A),(N,N)}S = \{(A, A), (A, N), (N, A), (N, N)\}
Each outcome is equally likely, with probability: P(each outcome)=14P(\text{each outcome}) = \frac{1}{4}
Now, define a random variable XX as the total number of bags that are adulterated.
Step-by-step:
Let’s find the value of XX for each sample point:
- (A,A)⇒X=2(A, A) \Rightarrow X = 2
- (A,N)⇒X=1(A, N) \Rightarrow X = 1
- (N,A)⇒X=1(N, A) \Rightarrow X = 1
- (N,N)⇒X=0(N, N) \Rightarrow X = 0
Now calculate the probability distribution of XX:
| XX | Values | Probability |
|---|---|---|
| 0 | (N, N) | 14\frac{1}{4} |
| 1 | (A, N), (N, A) | 14+14=12\frac{1}{4} + \frac{1}{4} = \frac{1}{2} |
| 2 | (A, A) | 14\frac{1}{4} |
Final Answer: Probability Distribution of XX
P(X=0)=14P(X=1)=12P(X=2)=14\begin{align*} P(X = 0) &= \frac{1}{4} \\ P(X = 1) &= \frac{1}{2} \\ P(X = 2) &= \frac{1}{4} \end{align*}
Explanation
This problem explores a fundamental concept in probability: defining a sample space, assigning probabilities, and determining the probability distribution of a random variable based on outcomes. In this case, we assume two bags of cement (one black and one white), each of which can independently be either adulterated (A) or not adulterated (N). The total number of combinations in the sample space is four, covering all possibilities: both adulterated, only one adulterated (either black or white), and neither adulterated.
Given that each outcome in the sample space is equally likely with a probability of 14\frac{1}{4}, we define a random variable XX as the total number of adulterated cement bags. This random variable can take on values 0, 1, or 2. We then determine the probability of each value of XX by counting how many outcomes in the sample space correspond to each value and summing their probabilities.
For instance, only one outcome—(N, N)—corresponds to 0 adulterated bags, so P(X=0)=14P(X = 0) = \frac{1}{4}. Two outcomes correspond to exactly one adulterated bag—(A, N) and (N, A)—so P(X=1)=12P(X = 1) = \frac{1}{2}. Finally, the outcome (A, A) yields two adulterated bags, so P(X=2)=14P(X = 2) = \frac{1}{4}.
This distribution is a discrete probability distribution and fully satisfies the condition that all probabilities are non-negative and sum up to 1. The concept illustrated here is foundational in both theoretical and applied probability, relevant in fields such as quality control, risk assessment, and reliability testing.
