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A geometric distribution deals with the number of trials needed to get the first success in a sequence of independent and identically distributed Bernoulli trials. In our exercise, finding the number of panels that need to be inspected before discovering a flaw follows this distribution. Here, a 'success' is finding at least one flaw in a panel, and the probability of this success happening on any given panel is determined by the Poisson distribution with a mean of 0.02 flaws per panel.
To find the expected value or the average number of panels to be inspected before finding a flaw, we use the formula for an expected value in a geometric distribution, which is:

• Expected number of panels = \( \frac{1}{p} \)

Where \( p \) is the probability of success on each trial, equivalent to finding at least one flaw. Since our success probability is derived from the rate of flaws, it is 0.02. Therefore, we calculate this as:

• Expected number of panels = \( \frac{1}{0.02} = 50 \)

So, on average, you would expect to inspect 50 panels before you find the first flaw.

The binomial distribution is useful when you perform a fixed number of independent experiments, each with two possible outcomes: success or failure. In this problem, conducting 50 independent inspections is modeled as a series of Bernoulli trials, with 'success' defined as finding a panel with at least one flaw.
For part (c) of our problem, we aim to find the probability that the number of panels with at least one flaw is 2 or fewer, using a binomial distribution. Here, the number of trials \( n \) is 50 (the panels), and the probability of success for each trial \( p \) is approximately 0.0198, derived from:

• Probability of finding at least one flaw = \( 1 - e^{-0.02} \)

The binomial probability formula which helps us calculate this is:

• \( P(Y = k) = \binom{n}{k} p^k (1-p)^{n-k} \)

To find \( P(Y \leq 2) \), calculate and sum \( P(Y = 0) \), \( P(Y = 1) \), and \( P(Y = 2) \). This gives us the total probability that 2 or fewer panels have flaws.

Probability calculations are crucial in statistics and allow us to infer and predict outcomes based on known probabilities. In this exercise, we use several probability distributions to compute specific probabilities.
Let's summarize the calculations made for each part of the exercise:

• Part (a): Calculating the probability using the Poisson distribution. Here, the probability that no flaws are found in 50 panels is given by \( P(X = 0) = \frac{e^{-1} \cdot 1^0}{0!} = e^{-1} \approx 0.3679 \).
• Part (b): Using the geometric distribution, the expected number of panels inspected before finding a flaw equals \( 50 \) panels.
• Part (c): Using the binomial distribution to find the probability that 2 or fewer panels have one or more flaws. This involves adding probabilities of having \( 0, 1, \) or \( 2 \) flawed panels.

These probability calculations help convert abstract events into tangible expectations, shedding light on the behavior of the process involving panel flaws.