Based on past experience, it is assumed that the number of flaws per metre in rolls of paper follows a Poisson distribution with a mean of 3 flaws per 10 metres of paper. What is the probability that more than 1 flaw is found in 5 metres of paper? What is the probability that there is more than a 3 metre distance between the first and second flaws?

The problem deals with the Poisson distribution and requires finding the probability of more than one flaw in 5 metres of paper and the probability of more than a 3-meter distance between the first two flaws. For the first part, one calculates the probabilities for 0 and 1 flaw and subtracts these from 1. For the second part, one calculates the probability in the exponential distribution.

Given that the mean (λ) is 3 flaws per 10 metres, for 5 metres of paper, λ would be 3/2=1.5 flaws. To find the probability of more than 1 flaw in 5 metres of paper, we first need to find the probabilities for 0 and 1 flaw and subtract these from 1.The formula to calculate these probabilities is P(x;λ)= e^-λ * λ^x / x!. For 0 flaws, x=0 and for 1 flaw, x=1. Therefore, 1- (P(0;1.5) + P(1;1.5)) gives the probability of more than one flaw in 5 metres of paper.

For the second question, when considering the distance between flaws in a Poisson process, the distribution to use is the exponential distribution. In this case, the expected distance between flaws is λ, which equals 10/3 meters for the given data. The probability that there is more than 3 meters between flaws is calculated as 1 - P(x<3) in the cumulative density function of the exponential distribution.

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