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A supplier of copper wire looks for flaws before despatching it to customers. It is known that the number of flaws follows a Poisson probability distribution with a mean of 2.3 flaws per metre. a) Determine the probability that there are exactly 2 flaws in 1 metre of the wire. b) Determine the probability that there is at least one flaw in 2 metres of the wire.

Short Answer

Expert verified
(a) The probability of exactly 2 flaws in 1 metre is approximately 0.227. (b) The probability of at least 1 flaw in 2 metres is approximately 0.990.

Step by step solution

01

Understanding the Problem

We have a Poisson distribution with a mean rate \( \lambda = 2.3 \) flaws per metre. We need to find two probabilities: (a) the probability of exactly 2 flaws in 1 metre, and (b) the probability of at least 1 flaw in 2 metres.
02

Using the Poisson Formula

The probability of observing \( k \) events in an interval for a Poisson distribution is given by the formula \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \). For part (a), set \( k = 2 \) and \( \lambda = 2.3 \).
03

Calculating for Exactly 2 Flaws

Substitute \( k = 2 \) and \( \lambda = 2.3 \) into the formula: \[ P(X = 2) = \frac{e^{-2.3} \times 2.3^2}{2!} \]. Calculate this to find the probability.
04

Using Complement Rule for At Least 1 Flaw

For part (b), first redefine the problem for a 2-metre length, which has a mean \( \lambda = 4.6 \) (since \( 2.3 \times 2 = 4.6 \)). Find the probability of having 0 flaws and use the complement rule: \[ P(\text{at least 1 flaw}) = 1 - P(X = 0) \].
05

Calculating for No Flaws in 2 Metres

Substitute \( k = 0 \) and \( \lambda = 4.6 \) into the Poisson formula: \[ P(X = 0) = \frac{e^{-4.6} \times 4.6^0}{0!} \]. Compute this probability.
06

Final Probability for At Least 1 Flaw

Subtract the probability of 0 flaws from 1 to find the probability of at least 1 flaw: \[ P(\text{at least 1 flaw}) = 1 - P(X = 0) \].

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Key Concepts

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

Probability Calculation
Probability calculation is a fundamental concept in statistics used to determine how likely an event is to occur. In the context of Poisson distribution, it helps us calculate the likelihood of a certain number of events happening within a fixed interval of time or space. The Poisson probability formula is used to find the probability of exactly - 'k' events occurring in a fixed interval. The distribution is defined by the equation: \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \), where:- \( e \) is the base of the natural logarithm, approximately 2.71828,- \( \lambda \) is the average number of occurrences in the interval,- \( k \) is the number of occurrences we want to find the probability for,- \( k! \) is the factorial of \( k \).For example, in the exercise, to find the probability of exactly 2 flaws in 1 metre of wire, we would plug \( k = 2 \) and \( \lambda = 2.3 \) into this formula. This calculation gives a precise probability of encountering that specific number of flaws in the wire.
Mean Rate
The mean rate, denoted as \( \lambda \), is a critical parameter in the Poisson distribution that represents the average rate at which events occur within a given interval. It tells us the typical number of occurrences we can expect.- In practical applications, like checking for flaws in copper wire, the mean rate allows us to assess the likelihood of different numbers of flaws.- For the problem provided, a mean rate of \( \lambda = 2.3 \) means that on average, 2.3 flaws are expected per metre of wire.- When the interval changes, such as checking over 2 metres of wire, the mean rate scales accordingly. For 2 metres, the mean rate is doubled, becoming \( \lambda = 4.6 \).Understanding the mean rate is invaluable for interpreting results and assessing risk. It allows you to adjust your expectations based on the length of wire inspected or any other varying parameter in similar scenarios.
Complement Rule
The complement rule is a helpful tool in probability when calculating the likelihood of at least one event occurring. It states that the probability of an event happening is 1 minus the probability of it not happening.- This is represented by: \( P(A) = 1 - P(\overline{A}) \), where \( P(A) \) is the probability of the event occurring, and \( P(\overline{A}) \) is the probability of it not occurring.- In the context of the given exercise, to find the probability of at least one flaw in 2 metres of wire, we first find the probability of having zero flaws using the Poisson formula.Once the solution for no flaws (\( k = 0 \)) is available, we simply subtract this probability from 1 to find the probability that there's at least one flaw. This simplifies calculations as it might be easier to calculate the probability of nothing happening instead of all possible events.

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