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!
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.
