Poisson Distribution Practice Questions

The Poisson distribution is a discrete probability distribution that gives the probability of a certain number of events occurring in a fixed interval of time, distance, area, or space, assuming the events occur randomly, independently, and at a constant average rate (λ).

Question 1: If 4% of the total items made by a factory are defective. Find the probability that less than 2 items are defective in the sample of 50 items.

Solution:  

Here we have, n = 50, p = (4/100) = 0.04, q = (1-p) = 0.96,  λ = 2

Using Poisson's Distribution,
P(X = 0) = \frac{2^0e^{-2}}{0!} = 1/e² = 0.13534
P(X = 1) = \frac{2^1e^{-2}}{1!} = 2/e² = 0.27068

Hence the probability that less than 2 items are defective in sample of 50 items is given by:
P( X > 2 ) = P( X = 0 ) + P( X = 1 ) = 0.13534 + 0.27068 = 0.40602

Question 2: If the probability of a bad reaction from medicine is 0.002, determine the chance that out of 1000 persons, more than 3 will suffer a bad reaction from medicine.

Solution:

Here we have, n = 1000, p = 0.002, λ = np = 2
X = Number of person suffer a bad reaction 

Using Poisson's Distribution

P(X > 3) = 1 - {P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)} 
P(X = 0) = \frac{2^0e^{-2}}{0!} = 1/e²
P(X = 1) =  \frac{2^1e^{-2}}{1!} = 2/e² 
P(X = 2) =  \frac{2^2e^{-2}}{2!} = 2/e²
P(X = 3) =  \frac{2^3e^{-2}}{3!} = 4/3e²
P(X > 3) = 1 - [19/3e²] = 1 - 0.85712 = 0.1428

Question 3: If 1% of the total screws made by a factory are defective. Find the probability that less than 3 screws are defective in a sample of 100 screws.

Solution:

Here we have, n = 100, p = 0.01, λ = np = 1
X = Number of defective screws

Using Poisson's Distribution

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) 
P(X = 0) =  \frac{1^0e^{-1}}{0!} = 1/e
P(X = 1) = \frac{1^1e^{-1}}{1!} =1/e
P(X = 2) = \frac{1^2e^{-1}}{2!} =1/2e

Thus, P(X < 3) = 1/e + 1/e +1/2e = 2.5/e = 0.919698

Question 4: If in an industry there is a chance that 5% of the employees will suffer from coronavirus. What is the probability that in a group of 20 employees, more than 3 employees will suffer from coronavirus?

Solution:

Here we have, n = 20, p = 0.05, λ = np = 1
X = Number of employees who will suffer corona

Using Poisson's Distribution

P(X > 3) = 1-[P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]
P(X = 0) = \frac{1^0e^{-1}}{0!} = 1/e
P(X = 1) = \frac{1^1e^{-1}}{1!} = 1/e
P(X = 2) =\frac{1^2e^{-1}}{2!} =1/2e
P(X = 3) =\frac{1^3e^{-1}}{3!} =1/6e
P(X > 3) = 1 - [1/e + 1/e + 1/2e + 1/6e]

⇒ P(X > 3) = 1 - [ 8/3e] = 0.018988

Practice Questions

Question 1: A customer support center receives an average of 8 calls per hour. Assuming the number of calls follows a Poisson distribution, find the probability that exactly 5 calls are received in the next hour.

Question 2: A factory produces electronic chips, and on average 3 defective chips are found in every batch of 200 chips. If defects follow a Poisson distribution, what is the probability that a randomly selected batch contains at most 2 defective chips?

Question 3: A hospital emergency department receives an average of 6 emergency patients every 30 minutes. Assuming arrivals follow a Poisson distribution, find the probability that more than 8 patients arrive in the next 30 minutes.

Question 4: A website experiences an average of 15 server errors per day. Assuming the number of errors follows a Poisson distribution, find the probability that during a particular day the website experiences between 10 and 18 errors (inclusive).

Question 5: A meteor shower produces an average of 2 meteors every 10 minutes. An astronomer observes the sky continuously for 1 hour.

1. Find the probability of observing exactly 15 meteors during the hour.
2. Find the probability of observing at least 12 meteors during the hour.
3. A customer support center receives an average of 8 calls per hour. Assuming the number of calls follows a Poisson distribution, find the probability that exactly 5 calls are received in the next hour.