Question

The number of flaws in a fiber optic cable follows a Poisson process with an average of 0.6 per 100 feet. a) Find the probability of exactly 2 flaws in a 200- foot cable. b) Find the probability of exactly 1 flaw in the first 100 feet and exactly 1 flaw in the second 100 feet.

Answer 1

Answer:

a) There is a 21.69% probability of exactly 2 flaws in a 200- foot cable.

b) There is a 10.84% probability of exactly 1 flaw in the first 100 feet and exactly 1 flaw in the second 100 feet.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

$e = 2.71828$ is the Euler number

$\mu$ is the mean in the given interval.

a) Find the probability of exactly 2 flaws in a 200- foot cable.

Average of 0.6 per 100 feet. So 1.2 per 200 feet.

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-1.2}*(1.2)^{2}}{(2)!} = 0.2169$

There is a 21.69% probability of exactly 2 flaws in a 200- foot cable.

b) Find the probability of exactly 1 flaw in the first 100 feet and exactly 1 flaw in the second 100 feet.

This is $P(X = 1)^{2}$

So

$P(X = 1) = \frac{e^{-0.6}*(0.6)^{1}}{(1)!} = 0.3293$

At each 100 feet, there is a 32.93% probability of exactly 1 flaw in each interval of 100 feet. So

$P = (0.3293)^{2} = 0.1084$

There is a 10.84% probability of exactly 1 flaw in the first 100 feet and exactly 1 flaw in the second 100 feet.