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.

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:

In which

x is the number of sucesses

is the Euler number

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.

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

So

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

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

its a little blurry tbh

step-by-step explanation:

where is abdul's work? don't see that anywhere posted : /

step-by-step explanation: