A manufacturer of LCD panels used in computer displays has decided to ship  only panels with at most one defective pixel. Defective pixels are the result of randomly located flaws in the glass used to make thepanel. To be profitable, the manufacturer must ship 90% of its panels. If an LCD panel has 1024 x 768 = 786 432 pixels, how small must the chance for a defective pixel be in order for the probability of shipping a panel to be 0.95?

To solve this problem, let's break it down step by step. The goal is to determine how small the probability of a defective pixel must be in order for a panel to have at most one defective pixel with a probability of 0.95.

### Step 1: Define the Problem Variables

- Let \( n \) be the total number of pixels in a panel: \( n = 1024 \times 768 = 786,432 \).
- Let \( p \) be the probability that a pixel is defective.
- We want the probability that a panel has at most one defective pixel to be at least 0.95.

### Step 2: Assume a Binomial Distribution

The number of defective pixels in a panel can be modeled by a binomial distribution:

\[
X \sim \text{Binomial}(n, p)
\]

The probability mass function of a binomial distribution is given by:

\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}
\]

Here, \( k \) represents the number of defective pixels.

### Step 3: Approximation Using Poisson Distribution

Since \( n \) is large and \( p \) is small, we can approximate the binomial distribution with a Poisson distribution. The Poisson distribution has a parameter \( \lambda = n \cdot p \), which is the expected number of defective pixels per panel.

The probability of finding exactly \( k \) defective pixels in a Poisson distribution is:

\[
P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}
\]

Thus, the probability of having at most one defective pixel is:

\[
P(X \leq 1) = P(X = 0) + P(X = 1)
\]

This can be written as:

\[
P(X \leq 1) = \frac{\lambda^0 e^{-\lambda}}{0!} + \frac{\lambda^1 e^{-\lambda}}{1!}
\]

Simplifying this:

\[
P(X \leq 1) = e^{-\lambda} + \lambda e^{-\lambda} = e^{-\lambda}(1 + \lambda)
\]

### Step 4: Set the Desired Probability

We want the probability \( P(X \leq 1) \) to be at least 0.95. Therefore:

\[
e^{-\lambda}(1 + \lambda) \geq 0.95
\]

Now, we need to solve this inequality for \( \lambda \), and recall that \( \lambda = n \cdot p \).

### Step 5: Solve for \( \lambda \)

To solve \( e^{-\lambda}(1 + \lambda) = 0.95 \), we can use numerical methods (e.g., trial and error, or a calculator) to find the approximate value of \( \lambda \).

Let me calculate this for you.The value of \( \lambda \), which satisfies the equation \( e^{-\lambda}(1 + \lambda) = 0.95 \), is approximately 0.355.

### Step 6: Relate \( \lambda \) to \( p \)

Recall that \( \lambda = n \cdot p \), where \( n = 786,432 \). Therefore, we can solve for \( p \):

\[
p = \frac{\lambda}{n}
\]

Now, let me calculate the value of \( p \).The probability \( p \) of a defective pixel must be approximately \( 4.52 \times 10^{-7} \) (or \( 0.000000452 \)) in order for the probability of shipping a panel (with at most one defective pixel) to be 0.95.

This means the chance of a defective pixel is extremely small, ensuring that most panels meet the shipping criteria.

Would you like more details or have any questions? Here are five related questions for further exploration:

1. How does changing the number of pixels in a panel affect the probability \( p \)?
2. What happens to the required probability \( p \) if the acceptable number of defective pixels increases to 2?
3. How would the solution differ if we did not approximate the binomial distribution with the Poisson distribution?
4. What is the expected number of defective panels given a specific probability \( p \)?
5. How does this manufacturing probability model relate to quality control in other industries?

**Tip:** When dealing with large numbers and small probabilities, using approximations like the Poisson distribution can significantly simplify complex calculations.

Calculate the probability of shipping LCD panels with at most one defective pixel, ensuring a 95% probability of shipment. Learn how to apply probability theory, binomial distribution, and Poisson distribution to solve this manufacturing problem.

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