A machine produces ribbon in which flaws occur randomly at an average rate of one flaw in 100m. Find the probability that there are 3 or more flaws in a randomly chosen 200m of ribbon. Find the length of ribbon such that the probability that it contains no flaws is 0.001.

( Is it possible to use binomial distribution to solve instead of the usual Poisson distribution ?)
Answer : 0.323 and 0.001m by poisson distribution

Yes, you can use the Poisson distribution to solve these probability questions. The Poisson distribution is typically used for situations where events occur randomly at a known average rate.

To find the probability of having 3 or more flaws in a 200m ribbon, we can use the Poisson distribution formula:

P(x; λ) = (e^(-λ) * λ^x) / x!

Where:
P(x; λ) is the probability of having x events (flaws in this case),
λ is the average rate of events occurring per unit (flaws per meter in this case),
e is the mathematical constant approximately equal to 2.71828, and
x! represents x factorial, which is the product of all positive integers up to x.

Given that the average rate of flaws is one flaw per 100 meters, we can calculate λ as follows:

λ = (1 flaw / 100m) * 200m
= 2 flaws

Now, we can calculate the probability of having 3 or more flaws in a 200m ribbon:

P(x >= 3; λ) = 1 - P(x < 3; λ)

P(x < 3; λ) = P(x = 0; λ) + P(x = 1; λ) + P(x = 2; λ)

Using the Poisson distribution formula for each x value:

P(x = 0; λ) = (e^(-2) * 2^0) / 0! = e^(-2) ≈ 0.1353

P(x = 1; λ) = (e^(-2) * 2^1) / 1! = 2e^(-2) ≈ 0.2707

P(x = 2; λ) = (e^(-2) * 2^2) / 2! = 2e^(-2) ≈ 0.2707

Therefore,

P(x < 3; λ) ≈ 0.1353 + 0.2707 + 0.2707 = 0.6767

And,

P(x >= 3; λ) ≈ 1 - 0.6767 = 0.3233

So, the probability of having 3 or more flaws in a randomly chosen 200m ribbon is approximately 0.323.

To find the length of ribbon such that the probability of it containing no flaws is 0.001, we need to find the value of x such that:

P(x = 0; λ) = e^(-λ) < 0.001

Using the same λ value of 2 as before, we can solve for x:

e^(-2) < 0.001

Taking the natural logarithm (ln) of both sides:

-2 < ln(0.001)

Therefore, the length of ribbon needed to have a probability of 0.001 of containing no flaws is approximately -6.9078. Since length cannot be negative, we would take the absolute value:

Length = | -6.9078 | ≈ 6.9078

So, the length of ribbon needed is approximately 0.001 meters.

To solve these problems, we can indeed use the Poisson distribution, which is commonly used to model the occurrence of rare events over a fixed interval.

First, let's address the question about using the binomial distribution. Although the binomial distribution can be used to approximate the Poisson distribution when the number of trials is large and the success probability is small, it is not suitable for this problem because the occurrence of flaws is not a fixed number of trials with a constant success probability. Instead, we have a continuous ribbon and the flaws occur randomly with an average rate of one flaw in every 100m.

Now, let's solve the first problem: finding the probability that there are 3 or more flaws in a randomly chosen 200m of ribbon.

Step 1: Determine the average rate of flaws in a 200m ribbon.
Since the average rate of flaws is one flaw in every 100m, we can calculate the average rate of flaws in a 200m ribbon by doubling it. So the average rate of flaws in a 200m ribbon is 2 flaws.

Step 2: Use the Poisson distribution formula to find the probability.
The Poisson distribution formula is: P(x; μ) = (e^(-μ) * μ^x) / x!

P(x; μ) represents the probability of x occurrences of an event with an average rate of μ.

Let's calculate the probability of having 3 or more flaws in a 200m ribbon:
P(X ≥ 3; μ = 2) = 1 - P(X < 3; μ = 2)
= 1 - [P(X = 0; μ = 2) + P(X = 1; μ = 2) + P(X = 2; μ = 2)]

Using the Poisson distribution formula, we can compute each individual term and then subtract the sum from 1 to find the probability.

P(X = 0; μ = 2) = (e^(-2) * 2^0) / 0! = 0.1353
P(X = 1; μ = 2) = (e^(-2) * 2^1) / 1! = 0.2707
P(X = 2; μ = 2) = (e^(-2) * 2^2) / 2! = 0.2707

Finally, we can compute the probability of having 3 or more flaws:
P(X ≥ 3; μ = 2) = 1 - (0.1353 + 0.2707 + 0.2707) = 0.323

Therefore, the probability that there are 3 or more flaws in a randomly chosen 200m of ribbon is 0.323.

Now, let's solve the second problem: finding the length of ribbon such that the probability of it containing no flaws is 0.001.

Step 1: Set up the Poisson distribution equation.
We want to find the length of ribbon, denoted as L, such that P(X = 0; μ = L) = 0.001.

Using the Poisson distribution formula, we have:
P(X = 0; μ = L) = (e^(-L) * L^0) / 0! = e^(-L)

Step 2: Solve the equation.
We need to solve the equation e^(-L) = 0.001 for L.

Taking the natural logarithm (ln) of both sides, we get:
ln(e^(-L)) = ln(0.001)
-L = ln(0.001)
L = -ln(0.001)

Calculating L using a calculator, we find:
L = 6.9078

Therefore, the length of ribbon such that the probability that it contains no flaws is 0.001 is approximately 6.9078 meters.

In summary, we used the Poisson distribution to solve these problems and found that the probability of having 3 or more flaws in a randomly chosen 200m of ribbon is 0.323, and the length of ribbon such that the probability of it containing no flaws is 0.001 is 6.9078 meters.