Very important Discrete Probability Distribution

1. Poisson Distribution
Very important Discrete Probability Distribution
Dr. Anjali Devi JS
Guest Faculty
School of Chemical Sciences
M G University

2. Poisson Distribution
Siméon Dens Poisson
Poisson- French word for Fish!!

3. Poisson Probability Distribution
Number of discrete events Given period of time
 A Poisson distribution is a tool that helps to predict the
probability of certain events from happening when you know
how often the event has occurred.
People Arriving In a minute
Phone calls
In a day
Cars passing through road
In an hour
 It give us the probability of a given number of events happening
in a fixed interval of time.
Examples

4. Condition for Poisson Distribution
Events are discrete (you can count them)
 Events cannot happen at the same time
 Events are independent

5. Poisson Parameter
λ
Lambda
The rate
Number of events per time period
Mean =λ
Variance =λ
Standard deviation = λ

6. Poisson Probability Distribution
Errors Per page of a book
Potholes
In a scoop of ice cream
Cookie dough lumps
In a km of road
Number of discrete events Given interval, distance or area

7. Notation
The random variable X
Is distributed
Using Poisson distribution with
parameter λ
Example Suppose X is a Binomial random variable with λ=4, then
notation:
𝑋~𝑃𝑜(4)
If a discrete random variable X has Poisson distribution (i.e., its probability
function uses a binomial distribution), then we write:
𝑋~𝑃𝑜(λ)

8. Formula
𝑃 𝑋 = 𝑥 =
𝑒−λλ𝑥
𝑥!
e=Euler’s constant= 2.718
λ=Mean
Poisson probability mass function,

9. A random variable X has a Poisson distribution with parameter λ such that P
(X = 1) = (0.2) P (X = 2). Find P (X = 0).
Question
𝑃 𝑋 = 𝑥 =
𝑒−λλ𝑥
𝑥!
𝑃 𝑋 = 1 =0.2[𝑃 𝑋 = 2 ]
𝑒−λλ1
1!
= 0.2 {
𝑒−λλ2
2!
}
λ=10
𝑃 𝑋 = 0 =
𝑒−10
λ0
0!
Answer
P (X= 0) = 0.0000454

10. One nanogram of Plutonium-239 will have an average of 2.3 radioactive
decays per second. The number of decays will follow a Poisson
distribution.
What is the probability that in 2second period there are exactly 3
radioactive decays?
Question
Let X represents number of decays in a 2 second period
λ = 2.3 𝑥2 = 4.6
𝑥 = 3
𝑃 𝑋 = 𝑥 =
𝑒−λ
λ𝑥
𝑥! P(X=3)=0.163

11. Event Formula for probability Probability mass function
0 𝑟𝑎𝑑𝑖𝑜𝑎𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑐𝑎𝑦 𝑖𝑛 𝑠𝑒𝑐𝑜𝑛𝑑
𝑋 = 0
𝑒−4.6
4.60
0!
0.010
𝑋 = 1 𝑒−4.6
4.61
1!
0.0462
𝑋 = 2 𝑒−4.64.62
2!
0.1064
𝑋 = 3 𝑒−4.64.63
3!
0.1631
𝑋 = 4 𝑒−4.64.64
4!
0.1876
𝑋 = 5 𝑒−4.6
4.65
5!
0.1726
𝑋 = 6 𝑒−4.6
4.66
6!
0.1323

12. 1/32
Visualising Poisson Distribution
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
X=0 X=1 X=2 X=3 X=4 X=5 X=6
Chart Title

13. Thank You