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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

Expert verified
a. Probability that there are two or more defective bulbs is approximately 0.00043. b. Expected number of automotive lights to check is approximately 2326.

Step by step solution

01

Define the Random Variable and Probability

Let the random variable \( X \) represent the number of defective bulbs in the array of 30 LED bulbs. We know that \( X \) follows a binomial distribution, \( X \sim \text{Binomial}(n=30, p=0.001) \).
02

Calculate the Complementary Probability

The problem asks for calculating the probability of two or more defective bulbs. We first find the complementary probability of having zero or one defective bulb: \( P(X=0) + P(X=1) \). The formula for the probability of \( k \) defectives is given by: \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]
03

Calculate \( P(X=0) \)

For zero defective bulbs (i.e., \( k=0 \)), we have: \[ P(X=0) = \binom{30}{0}(0.001)^0(0.999)^{30} = (0.999)^{30} \] Calculate \( (0.999)^{30} \) to get the probability for zero defective bulbs.
04

Calculate \( P(X=1) \)

For one defective bulb (i.e., \( k=1 \)), we have: \[ P(X=1) = \binom{30}{1}(0.001)^1(0.999)^{29} = 30 \times 0.001 \times (0.999)^{29} \] Calculate this to get the probability for one defective bulb.
05

Calculate \( P(X \geq 2) \)

Subtract the sum of the probabilities from Steps 3 and 4 from 1 to find the probability of two or more defective bulbs: \[ P(X \geq 2) = 1 - (P(X=0) + P(X=1)) \] Compute this value.
06

Use Geometric Distribution for Expected Number of Checks

To find the expected number of checks to get one automotive light with two or more defective bulbs, use a geometric distribution where the probability of success \( p \) is \( P(X \geq 2) \). The expected number \( E \) is given by: \[ E = \frac{1}{P(X \geq 2)} \] Compute \( E \) using the value from Step 5.

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Key Concepts

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

Probability
Probability is the measure of the likelihood that an event will occur. It ranges from 0 to 1, with 0 meaning an event will never happen, and 1 indicating it will definitely happen.
In the context of LED bulbs, we are interested in finding the probability that there are two or more defective bulbs. Since the probability of a single bulb being defective is 0.001, we consider this when calculating the probability associated with different numbers of defective bulbs.
  • A probability of 0 means no defective bulbs in our calculation.
  • A probability around 1.0 implies that it's almost certain to encounter defective bulbs.
When calculating the probabilities for different scenarios (like 0 or 1 defective bulb), we use the formula for binomial distribution, where we rely on multiplying combinations with probabilities raised to the power of occurrences.
Random Variable
A random variable is a numerical outcome of a random phenomenon. In our LED bulb example, we let a random variable, denoted as \( X \), represent the number of defective bulbs in an array of 30.
  • The type of random variable we're dealing with here is a discrete random variable, as it can take on specific integer values like 0, 1, 2, etc.
By defining \( X \), we can interpret different states of the world regarding the number of defective bulbs and use mathematical tools to deal with probabilities further. Random variables are essential as they allow the encapsulation of uncertain outcomes in a structured way. This helps when quantifying and comparing probabilities or calculating expected values later on.
Geometric Distribution
The geometric distribution is used when calculating the number of trials required for the first success in a series of independent and identically distributed Bernoulli trials. In our problem, we want to know how many automotive lights we need to check to find one with two or more defective bulbs.
  • Using the geometric distribution, we set our success criteria as finding a light with \( X \geq 2 \) defective bulbs.
  • The probability of this occurring, \( P(X \geq 2) \), is treated as our 'success' probability.
To compute the expected number of checks, we take the reciprocal of the probability of success. This effectively tells us how many trials we expect to conduct before the event occurs for the first time. This method uncovers the limitations and works under the assumption that each trial is independent, which in this case, the setup of our LED bulb check supports.
Expected Value
The expected value is a foundational concept in statistics representing the average outcome of a random variable that one anticipates. For the probability of bulbs, we calculate the expected number of checks needed to encounter a specific outcome.
  • In a geometric distribution, the expected value of the number of trials until the first success is \( E = \frac{1}{p} \), where \( p \) is the probability of achieving a success.
  • The expected value provides useful insight into how many automotive lights we would need to check on average to find one with two or more defective bulbs.
It doesn't predict the exact outcome in a single trial but gives a statistical measure of what to anticipate over many trials. Being aware of expected values is particularly useful for planning and managing expectations in probabilistic scenarios.

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