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Chapter 4: Problem 116

A rental truck agency services its vehicles on a regular basis, routinely checking for mechanical problems. Suppose that the agency has six moving vans, two of which need to have new brakes. During a routine check, the vans are tested one at a time. a. What is the probability that the last van with brake problems is the fourth van tested? b. What is the probability that no more than four vans need to be tested before both brake problems are detected? c. Given that one van with bad brakes is detected in the first two tests, what is the probability that the remaining van is found on the third or fourth test?

Short Answer

Expert verified
Based on the given problem, the probabilities we have calculated are as follows: a. The probability that the last van with brake problems is the fourth van tested: Probability a = 3(3!)/[(6!)/(2!)] b. The probability that no more than four vans need to be tested before both brake problems are detected: Probability b = ((4 C_2) * 4!)/[(6!)/(2!)] c. Given that one van with bad brakes is detected in the first two tests, the probability that the remaining van is found on the third or fourth test: Probability c = [(P(Case 1) * P(Case 1 or Case 2)) + (P(Case 2) * P(Case 1 or Case 2))]/(P(Case 1 or Case 2))

Step by step solution

01

Calculate the total possible test orderings

There are 6 vans, so there are 6! (6 factorial) ways to arrange the vans in order. However, we have two vans with faulty brakes, which are indistinguishable in terms of brake problems. So, we should divide this by 2! since both vans with brake problems could be interchanged while keeping the arrangement the same in terms of van types (non-faulty and faulty brakes). Therefore, the total number of van arrangements is: Total Orderings = \(\frac{6!}{2!}\)
02

Calculate the probability for part a

We need to find the number of arrangements where the fourth van tested has brake problems and is the last one to be discovered. That implies that the first three vans tested have one van with brake problems and two vans without. We can choose one of the first three spots for the van with brake problems in 3 different ways. Then, there are 3! ways to arrange the rest of the vans in the last three positions. So, the number of successful outcomes for part a is: Successful Outcomes a = 3 * 3! The probability for part a is: Probability a = \(\frac{\text{Successful Outcomes a}}{\text{Total Orderings}} = \frac{3*3!}{\frac{6!}{2!}}\)
03

Calculate the probability for part b

We need to find the number of arrangements where both vans with brake problems are detected within the first four vans tested. We can choose two of the first four spots for the brake problem vans in 4 choose 2 (\((4 C_2)\)) ways. Then, there are 4! ways to arrange the other four vans in the remaining positions. So, the number of successful outcomes for part b: Successful Outcomes b = \((4 C_2) * 4!\) The probability for part b is: Probability b = \(\frac{\text{Successful Outcomes b}}{\text{Total Orderings}} = \frac{(4 C_2) * 4!}{\frac{6!}{2!}}\)
04

Calculate the probability for part c

We are given that one of the vans with brake problems is found within the first two tests. There are two ways this can happen, the faulty van is either the first or the second tested. Let's calculate the probability for each case. Case 1: First van tested has brake problems Question: What is the probability that the second van with brake problems is found on the third or fourth test? Answer: There are three spots left, and one of them must have the second van with brake problems, so the probability is 2/3. Case 2: Second van tested has brake problems Question: What is the probability that the second van with brake problems is found on the third or fourth test? Answer: There are four spots left, and one of them must have the second van with brake problems, so the probability is 1/2. Now, we need to calculate the probability of having a van with brake problems in the first two tests. Probability of Case 1 and Case 2 is: P(Case 1 or Case 2) = \(\frac{(4 C_2) * 4! - (4 C_1) * 3! }{\frac{6!}{2!}}\) So, the probability for part c will be the weighted average of both cases. Probability c = \(\frac{P(Case 1) * P(Case 1 \text{ or } Case 2) + P(Case 2) * P(Case 1 \text{ or } Case 2)}{ P(Case 1 \text{ or } Case 2)}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is a branch of mathematics that deals with counting, arrangement, and combination of objects. It's an essential tool for calculating probabilities, especially in situations where you need to determine the number of possible outcomes. When you hear terms like permutations and combinations, you're dealing with combinatorics.
Let's dive into these concepts with an example. Imagine arranging six moving vans, out of which two have faulty brakes. To figure out all possible test orders, you'd calculate permutations, considering that the two faulty vans are indistinguishable within themselves. This means you'd calculate total arrangements as \(\frac{6!}{2!}\).

Why divide by \(2!\)? Because the two vans with brake problems are identical for the purpose of labeling them as faulty. By dividing by \(2!\), you adjust for these identical permutations, preventing them from inflating the count of unique arrangements that align with the scenario's needs.
  • **Permutations:** Arrangements where the order matters.
  • **Combinations:** Selections where the order does not matter.
In this exercise, permutations help determine the sequence of events, crucial when each test outcome is dependent on prior ones.
Statistical Inference
Statistical inference is the process of making educated guesses about a large group based on a sample. For example, when analyzing a repair shop testing a sample of moving vans to infer repair needs across their entire fleet. This involves using sample data to estimate probabilities and analyzes to comprehend patterns or tendencies.
In the rental truck scenario, statistical inference allows the agency to evaluate the likelihood of brake faulty between a small sample based on historical trends. Even from a few tests, conclusions about fleet-wide maintenance needs can be drawn. Though only dealing with six vans in this example, these principles scale to larger operations, allowing for strategic decision-making based on representative samples.

Statistical inference involves two main methods:
  • **Hypothesis Testing:** Decide if data support a certain hypothesis.
  • **Confidence Intervals:** Estimating a range of values for a population parameter based on a sample.
In practical terms, making decisions about servicing sequence, or predicting which van is most likely to have issues next, this example is a microcosm of statistical inference applied.
Conditional Probability
Conditional probability measures the probability of an event occurring given that another event has already occurred. It provides more precise probability estimates that account for specific circumstances or conditions.
In the context of the problem, we are asked to find the probability that a faulty van is detected at the third or fourth test given that one faulty van is already detected within the first two tests. This shifts the probability landscape, focusing only on remaining testcases because we've gained partial information already.

This can be calculated using the formula:
\[P(A|B) = \frac{P(A \cap B)}{P(B)}\]
Where:
  • **\(P(A|B)\):** Probability of event A given event B has happened.
  • **\(P(A \cap B)\):** Probability of both events A and B occurring together.
  • **\(P(B)\):** Probability of event B.
In this exercise, conditional probability enables decision-making based on past outcomes, illustrating how prior knowledge refines predictions and actions.

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Most popular questions from this chapter

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