Poisson distribution is a discrete probability distribution that describes the number of occurrences of events in a given interval of time or space.

Assumptions:

• The events occur independently of each other.
• The average number of occurrences in the interval is known and is given by the parameter λ.
• The number of occurrences in the interval is finite.

Probability mass function:

P(X = k) = (e^-λ) * λ^k / k!

where:

• P(X = k) is the probability of having exactly k occurrences.
• e is the base of the natural logarithm.
• λ is the average number of occurrences.
• k is the number of occurrences.
• k! is the factorial of k.

Moments:

• Mean (E) = λ
• Variance (Var) = λ
• Standard deviation (SD) = √λ

Applications:

• Counting the number of defects in a manufactured product.
• Predicting the number of calls received by a telephone in a given time interval.
• Modeling the number of accidents in a road traffic.

Example:

A factory produces 100 electronic components per day, and 2% of the components are defective. Assuming that the number of defective components is independent of the number of components produced in any given day, what is the probability that a randomly selected day will have exactly 3 defective components?

Using the Poisson distribution formula, the probability of having exactly k defective components is given by:

P(X = k) = (e^-λ) * λ^k / k!

where:

• P(X = k) is the probability of having exactly k defective components.
• λ is the average number of defective components per day, which is 2%.
• k is the number of defective components, which is 3.

Substituting these values into the formula, we get:

P(X = 3) = (e^-0.02) * 0.02^3 / 3! = 0.0039

Therefore, the probability of having exactly 3 defective components in a randomly selected day is 0.0039.