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Chapter 2: Problem 84

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities: a. \(P(\) all of the next three vehicles inspected pass) b. \(P\) (at least one of the next three inspected fails) c. \(P\) (exactly one of the next three inspected passes) d. \(P\) (at most one of the next three vehicles inspected passes) e. Given that at least one of the next three vehicles passes inspection, what is the probability that all three pass (a conditional probability)?

Short Answer

Expert verified
a. 0.343 b. 0.657 c. 0.189 d. 0.216 e. 0.3527

Step by step solution

01

Analyze and Setup the Problem

The problem involves calculating probabilities related to vehicles passing or failing an emissions inspection. Given: Probability of passing, \( P(\text{Pass}) = 0.7 \), and the probability of failure, \( P(\text{Fail}) = 0.3 \). Define a binomial probability distribution with \( n = 3 \), the number of vehicles inspected.
02

Calculate Probability for Part (a)

For (a), we want \( P(\text{all of the next three vehicles inspected pass}) = P(X = 3) \). Since the probability of passing is independent, \( P(X = 3) = (0.7)^3 \). Calculate: \[P(X = 3) = (0.7)^3 = 0.343.\]
03

Calculate Probability for Part (b)

Part (b) seeks \( P(\text{at least one fails}) = 1 - P(\text{all pass}) \). From Step 2, we have \( P(\text{all pass}) = 0.343 \). Therefore, \[P(\text{at least one fails}) = 1 - 0.343 = 0.657.\]
04

Calculate Probability for Part (c)

For (c), calculate \( P(\text{exactly one passes}) = P(X = 1) \). Using the binomial formula, \[P(X = 1) = \binom{3}{1} (0.7)^1(0.3)^2 = 3 \times 0.7 \times 0.09 = 0.189.\]
05

Calculate Probability for Part (d)

For (d), \( P(\text{at most one passes}) = P(X = 0) + P(X = 1) \). First, calculate \( P(X = 0) \):\[P(X = 0) = (0.3)^3 = 0.027.\]Thus, \[P(\text{at most one passes}) = 0.027 + 0.189 = 0.216.\]
06

Calculate Conditional Probability for Part (e)

For (e), find \( P(\text{all three pass} \mid \text{at least one passes}) \). This conditional probability is given by \[P(A|B) = \frac{P(A \cap B)}{P(B)}\]where, \( A: \text{all three pass}, \ B: \text{at least one passes} \). Since \( A \cap B = A \), simply use:\[P(A|B) = \frac{P(A)}{1 - P(X = 0)} = \frac{0.343}{1 - 0.027} = 0.3527.\]

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Key Concepts

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

Binomial Distribution
The binomial distribution is a probability distribution used to model the number of successes in a fixed number of independent Bernoulli trials. Each trial can result in just two outcomes, commonly termed as "success" or "failure". By following this problem, we set up a binomial distribution with 3 trials (as 3 vehicles are inspected). The probability of success (vehicle passing the inspection) is given as 0.7, and thus the probability of failure is 0.3.
  • The formula for the binomial probability of having exactly k successes in n trials is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where:
    • \( \binom{n}{k} \) is the number of ways to choose k successes from n trials.
    • p is the probability of success for each trial.
In our example, we calculate the probabilities for different scenarios like 0, 1, or all vehicles passing based on these principles.
Conditional Probability
Conditional probability is a measure of the probability of an event occurring given that another event has occurred. In our exercise, we're interested in the probability of all three vehicles passing the inspection given that at least one passes. This is expressed with the formula:
  • \( P(A|B) = \frac{P(A \cap B)}{P(B)} \)
Here, event A is "all three pass", and event B is "at least one passes". Importantly, since if all vehicles pass, at least one also logically passes, we have that \( P(A \cap B) = P(A) \), simplifying our calculations.
For this specific exercise, we calculated:
  • Cumulative probability of at least one vehicle passing, \( P(B) = 1 - P(X=0) \).
  • Thus, \( P(A|B) = \frac{0.343}{1 - 0.027} = 0.3527 \).
    • This kind of probability helps in understanding dependencies between events.
    Probability Calculation
    Probability calculation refers to the process of determining the likelihood of various outcomes. It's the backbone of understanding the binomial distribution and events like conditional probability. Through this lens, all specific probabilities in the problem, like the probability of exactly one vehicle out of three passing, are calculated using formulas.
    • For example, the probability of exactly one vehicle passing is:
      \[ P(X = 1) = \binom{3}{1} (0.7)(0.3)^2 = 0.189 \]
      This probability calculation relies heavily on understanding both factorial calculations for combination choices and powers for probability distributions.
    These detailed calculations form the essence of probability theory and require understanding both core formulas and the contexts they are applied in.
    Statistical Independence
    Statistical independence implies that the outcome of one event does not affect the outcome of another. In this context, knowing whether one vehicle passes the inspection gives zero information on whether the next vehicle will pass. Thus, we can multiply the independent probabilities together to find outcomes for multiple events. Each vehicle's result is independent of the others.
    • For instance, calculating all three vehicles passing we use:
      \[ P(X=3) = (0.7)^3 = 0.343 \]
      This multiplication across separate trials relies on statistical independence.
    By comprehending statistical independence, one can simplify real-world scenarios into manageable probability problems, aiding both in problem-solving and understanding deeper statistical principles.

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    Most popular questions from this chapter

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