3-176. Flaws occur in the interior of plastic used for automobiles according to a Poisson distribution with a mean of0.02 flaw per panel.(a) If 50 panels are inspected, what is the probability thatthere are no flaws?(b) What is the expected number of panels that need to beinspected before a flaw is found?(c) If 50 panels are inspected, what is the probability that thenumber of panels that have one or more flaws is less than orequal to 2?

To solve these problems, we'll use the properties of the Poisson distribution. The mean number of flaws per panel is given as \( \lambda = 0.02 \).

### (a) Probability of No Flaws in 50 Panels

For 50 panels, the mean number of flaws is:

\[
\lambda_{50} = 50 \times 0.02 = 1
\]

The probability of observing \( k \) flaws in a Poisson distribution is given by:

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

For \( k = 0 \):

\[
P(X = 0) = \frac{e^{-1} \cdot 1^0}{0!} = e^{-1} \approx 0.3679
\]

So, the probability of no flaws in 50 panels is approximately **0.3679**.

### (b) Expected Number of Panels Inspected Before Finding a Flaw

The number of panels that need to be inspected until the first flaw is found follows a geometric distribution. The expected value for a geometric distribution with success probability \( p \) is given by:

\[
E(X) = \frac{1}{p}
\]

The probability \( p \) of finding at least one flaw in one panel is:

\[
p = 1 - P(X = 0) = 1 - e^{-0.02} \approx 0.0198
\]

Thus, the expected number of panels inspected before finding a flaw is:

\[
E(X) = \frac{1}{0.0198} \approx 50.51
\]

### (c) Probability that the Number of Panels with One or More Flaws is ≤ 2

The number of panels with one or more flaws in 50 panels follows a binomial distribution, where the probability of success (finding at least one flaw) in one panel is \( p \).

Using the probability of having at least one flaw in a panel:

\[
p = 1 - e^{-0.02} \approx 0.0198
\]

Let \( Y \) be the number of panels with one or more flaws. We can model this as \( Y \sim \text{Binomial}(n=50, p) \).

To find \( P(Y \leq 2) \):

\[
P(Y \leq 2) = P(Y = 0) + P(Y = 1) + P(Y = 2)
\]

Calculating each term:

1. \( P(Y = 0) = (1 - p)^{50} \approx (0.9802)^{50} \approx 0.3679 \)
2. \( P(Y = 1) = \binom{50}{1} p (1 - p)^{49} \approx 50 \cdot 0.0198 \cdot (0.9802)^{49} \approx 0.3679 \)
3. \( P(Y = 2) = \binom{50}{2} p^2 (1 - p)^{48} \approx \frac{50 \cdot 49}{2} \cdot (0.0198)^2 \cdot (0.9802)^{48} \)

Combining these probabilities, we can compute \( P(Y \leq 2) \).

In summary, the calculations yield:

- **(a)** Probability of no flaws: **0.3679**
- **(b)** Expected panels until a flaw: **50.51**
- **(c)** Probability of ≤ 2 panels with flaws: calculate \( P(Y \leq 2) \).

Would you like to see detailed calculations for part (c), or do you have any other questions? Here are some related questions to consider:

1. What would the probabilities look like if the mean number of flaws per panel increased?
2. How does the expected number of panels inspected change with a different flaw rate?
3. Can you explain how the Poisson distribution is derived?
4. What real-world factors could affect the assumption of a Poisson distribution in this context?
5. How would you approach a scenario where flaws occur according to a different distribution? 

**Tip:** When dealing with probabilities, always double-check the calculations of factorial and powers, especially in combinatorial contexts.

3-176. Flaws occur in the interior of plastic used for automobiles according to a Poisson distribution with a mean of 0.02 flaw per panel. (a) If 50 panels are inspected, what is the probability that there are no flaws? (b) What is the expected number of panels that need to be inspected before a flaw is found? (c) If 50 panels are inspected, what is the probability that the number of panels that have one or more flaws is less than or equal to 2?

Explore the probability of flaws in 50 automobile plastic panels using Poisson distribution. This analysis covers the likelihood of no flaws, expected panels inspected before a flaw, and the probability of finding two or fewer flawed panels. Ideal for high school students studying probability.

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