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An array of 30 LED bulbs is used in an automotive light. The probability that a bulb is defective is 0.001 and defective bulbs occur independently. Determine the following: (a) Probability that an automotive light has two or more defective bulbs. (b) Expected number of automotive lights to check to obtain one with two or more defective bulbs.

Short Answer

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(a) Probability of two or more defective bulbs is about 0.0005. (b) Expect to check about 2000 automotive lights to find one with two or more defective bulbs.

Step by step solution

01

Understand the Probability Model

We have an array of 30 LED bulbs with each bulb having a defect probability of 0.001. Defective bulbs occur independently, which suggests the binomial probability distribution fits this situation. In binomial terms, we have \( n = 30 \) trials and \( p = 0.001 \) probability of success (defect).
02

Calculate the Probability of Two or More Defects

To find the probability of two or more defective bulbs, first calculate the probability of 0 and 1 defective bulb. Use the binomial probability formula: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \). Calculate \( P(X = 0) \) and \( P(X = 1) \), then find \( P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \).
03

Compute Probability of Zero Defects

For 0 defects (\( k=0 \)): \[ P(X = 0) = \binom{30}{0} \times 0.001^0 \times (1 - 0.001)^{30} = 0.9704 \]
04

Compute Probability of One Defect

For 1 defect (\( k=1 \)): \[ P(X = 1) = \binom{30}{1} \times 0.001^1 \times (1 - 0.001)^{29} \approx 0.0291 \]
05

Calculate Combined Probability for Two or More Defects

Now compute \( P(X \geq 2) = 1 - P(X = 0) - P(X = 1) = 1 - 0.9704 - 0.0291 \approx 0.0005 \). This is the probability of having two or more defective bulbs.
06

Calculate the Expected Number of Trials to Obtain Two or More Defects

Since the probability of two or more defects is \( 0.0005 \), use the expected value formula for geometric distribution: \( E(n) = \frac{1}{P(X \geq 2)} = \frac{1}{0.0005} = 2000 \). Therefore, you would expect to check 2000 lights to find one with two or more defects.

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Key Concepts

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

Probability of Defects
In an array of LED bulbs, each having a small chance of being defective, we want to figure out the probability of encountering defective bulbs. Imagine a box of 30 bulbs, where each one has a 0.1% chance of being bad. This tiny chance of defect is important because it helps us calculate the likelihood of multiple defective bulbs in the same setup.

To determine the probability of getting a certain number of defects, we use the binomial probability formula. This lets us calculate how likely it is to have 0, 1, or more defective bulbs in our group of 30. For instance, knowing the probabilities of 0 and 1 defective bulb allows us to find out the chance of getting two or more defective ones, which might be especially important when assessing product reliability.
Independent Trials
When mentioning that defective bulbs occur independently, it suggests a very specific mathematical outcome. Each bulb's chance of being defective does not affect another's. This means whether one bulb is defective or not, it doesn't change the probability of the next one being defective.

In probability terms, this is crucial. It allows using the binomial distribution accurately, as this distribution assumes trials are independent. Simply put, each bulb is its own event. This independence ensures calculations regarding defect probabilities are straightforward since each event remains unaffected by previous results.
Geometric Distribution
In scenarios where repeated trials are needed until a specific outcome occurs, the geometric distribution becomes relevant. Like when we want to know how many automotive light checks, on average, are necessary until we find one with two or more defective bulbs.

The key here is using the geometric distribution to figure out expected outcomes over repeated independent trials. The geometric distribution helps calculate the expected number of trials, working well when a certain event (like finding defects) is rare. It allows for understanding the average number of trials needed until a specific outcome (two or more defects) is observed.
Expected Value Calculation
Expected value is an essential concept in probability, guiding us to anticipate outcomes over the long term. It's like predicting the average result if an experiment were repeated many times. For example, if you check lights frequently, you want to know how many you must check before finding one with two or more defects.

When utilizing the geometric distribution, the expected value calculation becomes: \[ E(n) = \frac{1}{P(X \geq 2)} \]This formula indicates that, on average, you'll need 2000 automotive light checks to find one with two or more defective bulbs. It provides crucial insight into planning checks and understanding how frequently a specific problem, like multiple defects, appears.

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