This document discusses probability distributions and introduces the binomial and Poisson distributions. It begins with objectives of understanding probability distributions and applying them to public health challenges. It then defines probability distributions and covers discrete distributions, focusing on the binomial distribution which models success/failure trials and the Poisson distribution which models rare events over time. Examples are provided to demonstrate calculating probabilities using these distributions.

1. Probability Distributions
Lecturer: Aladejebi Oluwatoyin

2. OUTLINE
• Objectives
• Introduction
• Concept of Probability distribution
• Discrete Probability Distribution
– Binomial distribution
– Poisson distribution
• Conclusion

3. OBJECTIVES
• At the end of this lecture students will have an
understanding of:
• The notion and types of probability distributions
• The concept of the Binomial and Poisson
distributions
• How to calculate the Binomial and Poisson
distributions
• How to apply the knowledge to interpret and solve
public health challenges

4. INTRODUCTION
• Everyday, we are faced with the task of making
decisions amidst uncertainties.
• Many health issues ranging from behavioural change
to health choices are shrewd with uncertainties and
possibility of several outcomes that affect decision
making.
• As we have seen in previous lectures. This is where
“Probability” a process which calculates the possible
outcomes of given events together with the outcomes’
relative likelihoods and distribution comes in.

5. INTRODUCTION
• Probability distributions bring probability theory into
practical use.
• Thereby enabling us understand how these possible
outcomes of a random variable are linked with the
likelihood of occurrence.

6. CONCEPT OF PROBABILITY DISTRIBUTION
• To understand the concept of probability distribution,
there’s need to understand that Random Variables are
variables whose values are as a result of the outcome
of a statistical experiment/study.
• A probability distribution is defined as a list of all of
the possible outcomes of a random variable along with
their corresponding probability values.
• It can be expressed as a table, graph or an equation
that links each possible value that a random variable
can assume with its probability of occurrence.

7. CONCEPT OF PROBABILITY DISTRIBUTION
• For example, The exercise of flipping a coin two times.
can have four possible outcomes: HH, HT, TH, and TT.
• Probability distributions can be identified as being a
discrete probability distribution or continuous
probability distribution.

8. DISCRETE PROBABILITY DISTRIBUTION
• Discrete Probability Distribution -This is a
distribution which describes the probability of
occurrence of each value of a discrete random
variable.
• In the previous example given on the exercise of
flipping a coin twice with four possible outcomes: HH,
HT, TH, and TT
• We use the variable X to represent the number of
heads that result from the coin flips. The variable X
can take on the values 0, 1, or 2; thus making X a
discrete random variable.

9. DISCRETE PROBABILITY DISTRIBUTION
• The table below shows the probabilities associated
with each possible value of X. The probability of
getting 0 heads is 0.25; 1 head, 0.50; and 2 heads,
0.25. Thus, making the table an example of a
probability distribution for a discrete random
variable.
Number of heads, x Probability, P(x)
0 0.25
1 0.50
2 0.25

10. DISCRETE PROBABILITY DISTRIBUTION
• For the purpose of the lecture, we’ll be considering
two types of discrete probability distribution. They
are;
– Binomial distribution
– Poisson distribution
• Binomial Distribution – This is a discrete probability
distribution in which there are two outcomes to a
trial (success of probability ‘p’ or failure ‘q’), which is
repeated ‘n’ times with each repetition being
independent.
• Therefore, if the probability of a success is (p), the
probability of failure is 1-p or q because p+q =1.

11. DISCRETE PROBABILITY DISTRIBUTION
• This is expressed mathematically as
• Pr(r) =
𝑛!
𝑛−𝑟 !𝑟!
pr (1-p)𝑛-r
• Where r is the number of success in the trials
• Where n is the number of trials
• Where p is the probability of success in each trial
Example
• It was observed in a health facility that when tetanus
affects newborn infants only 10% recover. In a
random sample of 5 affected newborns. What is the
probability that two such affected newborns will

12. DISCRETE PROBABILITY DISTRIBUTION
• Solution
Probability for two to recover will be
Given n = 5, r = 2
Pr(2) =
5!
5−2 !2!
(0.1)2 (1-0.1)5-2
n! = n (n-1)(n-2)....
Therefore
Pr(2) =
5𝑋4𝑋3𝑋2𝑋1
3𝑋2𝑋1 𝑋 (2𝑋1)
(0.1)2 (0.9)3
Pr(2) = 0.0729 or 7.3%

13. DISCRETE PROBABILITY DISTRIBUTION
• Classwork
What is the probability that
• i. None of such newborns will recover i.e. r = 0
• ii. At least four such newborns will recover i.e. r = 4 or
5

14. DISCRETE PROBABILITY DISTRIBUTION
• Poisson Distribution – This is a discrete probability
distribution of a number of events occurring in a fixed
period of time if these occur with known average rate
and are independent of the time since the last event.
• It occurs when there are events which do not occur
as outcomes of a definite number of trials but at
random points of time and space.
• It is sometimes referred to as an approximation to the
Binomial distribution when the probability of a
success ‘p’ is small and the number of trials ‘n’ is
large.

15. DISCRETE PROBABILITY DISTRIBUTION
• That is, ‘n’ the number of trials indefinitely large n ∞
• ‘p’ the constant probability of success for each trial
and is definitely small, that is given as p 0
• It is mathematically expressed as
Pr(r) =
𝑒−μ
μr
𝑟!
Where e (a constant) = 2.718,
r! = (r)(r-1)(r-2).....
μ = mean value (may be known or estimated by the
sample arithmetic mean)

16. DISCRETE PROBABILITY DISTRIBUTION
Example
• The number of deaths from neonatal tetanus cases
presenting in a clinic averages 4 per year. Assuming a
Poisson distribution is appropriate. What will be the
probability of 6 deaths due to neonatal tetanus cases
presenting at same clinic yearly;
• Solution
• r= 6
• μ= 4
• Pr(6) =
𝑒−4
0.0046
6!

17. DISCRETE PROBABILITY DISTRIBUTION
• Pr(6) =
𝑒−4
46
6𝑋5𝑋4𝑋3𝑋2𝑋1
• Pr(6) =
0.0183 x4096
6𝑋5𝑋4𝑋3𝑋2𝑋1
• Pr(6) =
74.9568
720
• = 0.10 or 10%

18. DISCRETE PROBABILITY DISTRIBUTION
• Classwork
In a large survey of 100,000 births in country A. It was
observed that the incidence rate of a known fatal
condition is 412 per 100,000. In a random sample of 50
births from this population. What is the probability that
• i. No fatal case is found
• ii. There were two or more fatal cases.

19. Thank you!