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Chapter 4: Problem 83

An electronic office product contains 5000 electronic components. Assume that the probability that each component operates without failure during the useful life of the product is \(0.999,\) and assume that the components fail independently. Approximate the probability that 10 or more of the original 5000 components fail during the useful life of the product

Short Answer

Expert verified
Use a Poisson distribution with \(\lambda = 5\) to find \(P(X \geq 10)\). Approximate this probability by calculating \(1 - P(X < 10)\).

Step by step solution

01

Understanding the Problem

We have 5000 components, each with a probability of operating without failure of 0.999. We want to find the probability that 10 or more components fail.
02

Identifying the Type of Distribution

Since we are dealing with a large number of trials (5000 components), and the probability of success (each component operating without failure) is close to 1, we can approximate using a Poisson distribution.
03

Setting Up the Poisson Distribution Parameters

The parameter for the Poisson distribution, \(\lambda\), is the mean number of failures. It is given by \(n \cdot p_f\), where \(p_f = 1 - p\). Thus, \(\lambda = 5000 \times 0.001 = 5\).
04

Applying the Poisson Distribution

We want to find \(P(X \geq 10)\) for the Poisson distribution, where \(X\) is the number of failures. This is equivalent to \(1 - P(X < 10)\).
05

Calculating the Cumulative Probability

Using the Poisson Cumulative Distribution Function, calculate \(P(X < 10)\). This means we sum up \(P(X = k)\) for \(k = 0\) to \(9\) using \(P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\).
06

Finding the Complementary Probability

Subtract \(P(X < 10)\) from 1 to find \(P(X \geq 10)\). This gives us the probability that 10 or more components fail.

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Key Concepts

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

Probability of Failure
The probability of failure in any given situation tells us how likely it is that an event won't go as planned. In this particular exercise, each component of the electronic office product has a small chance, denoted as 0.001, of failing. This is derived from the probability of success, which is 0.999, meaning each component works without issues most of the time.

In probabilistic terms, failure probability is calculated as:
  • Failure Probability, \( p_f = 1 - p \)
  • Where \( p \) is the success probability (here 0.999)
Understanding these probabilities helps us predict outcomes in complex systems and assess potential risks. In larger systems, like with 5000 components, small probabilities of failure can still lead to notable failure events.
Independent Events
In probability, when we say events are independent, we mean that the occurrence of one does not affect the other. This is important when analyzing systems with many components.

Assuming each component in our exercise can fail independently means the failure of one does not increase or decrease the chance of another failing. This independence is crucial in calculating probabilities using the Poisson distribution.
  • Independence allows us to treat each failure separately.
  • This simplifies the calculation of the overall probability significantly.
For 5000 independent components, their collective behavior can be more easily approximated with statistical models like the Poisson distribution.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) is used to describe the probability of a random variable taking a value less than or equal to a certain level. In this problem, we're interested in the CDF of a Poisson distribution.

When calculating the probability that fewer than 10 components fail, we use the CDF:
  • We sum probabilities for all counts from 0 up to 9.
  • This involves using the Poisson probability mass function, \( P(X = k) = \frac{e^{- ext{位}} ext{位}^k}{k!} \)
To find the chance that 10 or more components fail, we take the complement, \( 1 - P(X < 10) \), reflecting its utility in understanding the total behavior of random variables up to a certain point.
Approximation Methods
Approximation methods are crucial when dealing with complex probability problems. Here, the use of a Poisson distribution makes problem-solving more feasible for calculating the likelihood of multiple failures among many components.

Why do we approximate with Poisson?
  • When the number of trials is large (like 5000) and the probability of each event is low, calculations directly using a binomial distribution become cumbersome.
  • The Poisson distribution provides a simpler calculation by using a single parameter, \( ext{位} \), where \( ext{位} = np_f \).
This method allows for practical and efficient estimation of probabilities in systems where direct calculation would otherwise be resource-intensive.

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Most popular questions from this chapter

Suppose the length of stay (in hours) at an emergency department is modeled with a lognormal random variable \(X\) with \(\theta=1.5\) and \(\omega=0.4\). Determine the following: (a) mean and variance (b) \(P(X<8)\) (c) Comment on the difference between the probability \(P(X<0)\) calculated from this lognormal distribution and a normal distribution with the same mean and variance.

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