Probability Distributions Guide for Decision Making

Probability Distributions
Venkata Sai Krishna M

What is Probability Distribution?
• How the probability is distributed among all possible outcomes
First Toss Second Toss Number of Tails Probability of the outcomes
Tails Tails 2 =0.5 * 0.5 = 0.25
Heads Tails 1 =0.5 * 0.5 = 0.25
Tails Heads 1 =0.5 * 0.5 = 0.25
Heads Heads 0 =0.5 * 0.5 = 0.25
Number of Tails First Toss Probability of the outcomes
0 (Heads,Heads) 0.25
1 (Heads,Tails); (Tails,Heads) =0.25+025=0.5
2 (Tails,Tails) 0.25
0
0.1
0.2
0.3
0.4
0.5
0.6
0 1 2
Probability of the outcomes

Types of Probability Distributions
• Discrete Probability Distributions
• Continuous Probability Distributions

Discrete Probability Distributions
• Can take only limited number of values
• Limited set of possible outcomes and nothing apart from that

Continuous Probability Distributions
• The considered variable can take any value from the given range
• A Convenient way to represent discrete distributions that have many
possible outcomes

Check!!!!
a) A probability provides information about the long run or expected
frequency of each outcome of an experiment
b) The graph of a probability distribution has the possible outcomes of
an experiment marked on the horizontal axis
c) A probability distribution lists the probabilities that each outcome is
random
d) A distribution is always constructed from a set of observed
frequencies like a frequency distribution
e) A probability distribution may be based on subjective estimates of
the likelihood of certain outcomes

Random Variables
• A variable is random if it takes different values as a result of outcomes
of random experiment
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115
Probability of a random variable
Patients in a day
No. of Days the
count is observed
Probability of a
random variable
100 1 0.01
101 2 0.02
102 3 0.03
103 5 0.05
104 6 0.06
105 7 0.07
106 9 0.09
107 10 0.10
108 12 0.12
109 11 0.11
110 9 0.09
111 8 0.08
112 6 0.06
113 5 0.05
114 4 0.04
115 2 0.02
Total 100 1.00

Expected value of random variable
Patients in a day
No. of Days the
count is observed
Probability of a
random variable
Patients in a day *
Probability of a random
variable
100 1 0.01 1.00
101 2 0.02 2.02
102 3 0.03 3.06
103 5 0.05 5.15
104 6 0.06 6.24
105 7 0.07 7.35
106 9 0.09 9.54
107 10 0.10 10.70
108 12 0.12 12.96
109 11 0.11 11.99
110 9 0.09 9.90
111 8 0.08 8.88
112 6 0.06 6.72
113 5 0.05 5.65
114 4 0.04 4.56
115 2 0.02 2.30
Expected Value: 108.02
The hospital should be prepared for
108.2 number of patients in a day.

Problem 1
Calculate the expected value based on the following frequency
distribution
Outcome Frequency
2 24
4 22
6 16
8 12
10 7
12 3
15 1

Problem 1
Calculate the expected value based on the following frequency
distribution
Outcome Frequency
Probability of each
outcome
Expected
Probability
2 24 0.28 0.56
4 22 0.26 1.04
6 16 0.19 1.13
8 12 0.14 1.13
10 7 0.08 0.82
12 3 0.04 0.42
15 1 0.01 0.18
Expected Value: 5.28

Problem 2
Jim purchased a video camera at a store of cost $300. The store manager offered a
service warranty of 5 years with an extra amount of $100. Jim did a some research
on maintenance expenses for the video camera and found out the probabilities as
follows. Based on the information please suggest taking warranty of worth $100 is
recommended?
Expense 0 50 100 150 200 250 300
Proabability 35 0.25 0.15 0.1 0.08 0.05 0.02

Problem 2
Jim purchased a video camera at a store of cost $300. The store manager offered a
service warranty of 5 years with an extra amount of $100. Jim did a some research
on maintenance expenses for the video camera and found out the probabilities as
follows. Based on the information please suggest taking warranty of worth $100 is
recommended?
Expense 0 50 100 150 200 250 300
Proabability 0.35 0.25 0.15 0.1 0.08 0.05 0.02
Expected Probability 0 12.5 15 15 16 12.5 6
Expected Value 77

Problem 3
A fire marshal in Maryland compiling a report of fire accidents from
past 2 years to get the expected fire accidents in a month. And he also
wants to identify fire accidents during winter. Please help him.
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
2015 25 30 15 10 10 5 2 2 1 4 8 10
2016 20 25 10 8 5 2 4 0 5 8 10 15

Problem 3
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
2015 25 30 15 10 10 5 2 2 1 4 8 10
2016 20 25 10 8 5 2 4 0 5 8 10 15
Number of
fire accidents
Frequency Probability
Expected
probability
0 1 0.0417 0.00
1 1 0.0417 0.04
2 3 0.1250 0.25
4 2 0.0833 0.33
5 3 0.1250 0.63
8 3 0.1250 1.00
10 5 0.2083 2.08
15 2 0.0833 1.25
20 1 0.0417 0.83
25 2 0.0833 2.08
30 1 0.0417 1.25
Expected Value for year 9.75
Number of fire
accidents
Freque
ncy
Probability
Expected
probability
10 1 0.1667 1.67
15 1 0.1667 2.50
20 1 0.1667 3.33
25 2 0.3333 8.33
30 1 0.1667 5.00
Expected Value
for winter
20.83

Expected value in decision making
• How the expected value is being used for monetary gain or loss?
• Types of losses that can be figured out:
• Obsolescence Losses:
• Having too much stock but no sales
• Opportunity Losses
• Having less stock and more sales

Sample Problem for explanation
A fruit store manager:
Daily Sales
Number of
Days Sold
Probability of
each number
being sold
10 15 0.15
11 20 0.20
12 40 0.40
13 25 0.25
Possible Stock Options
Possible
Requests
10 11 12 13
10 ₹ - ₹ 20 ₹ 40 ₹ 60
11 ₹ 30 ₹ - ₹ 20 ₹ 40
12 ₹ 60 ₹ 30 ₹ - ₹ 20
13 ₹ 90 ₹ 60 ₹ 30 ₹ -

Loss Calculation
Daily Sales Option 1
Probability of
each number
being sold
Expected
Loss
10 0 0.15 0.00
11 30 0.20 6.00
12 60 0.40 24.00
13 90 0.25 22.50
Total Loss 52.50
Probability of
each number
being sold
Expected
Loss
10 20 0.15 3.00
11 0 0.20 0.00
12 30 0.40 12.00
13 60 0.25 15.00
Total Loss 30.00
Probability of
each number
being sold
Expected
Loss
10 40 0.15 6.00
11 20 0.20 4.00
12 0 0.40 0.00
13 30 0.25 7.50
Total Loss 17.50
Probability of
each number
being sold
Expected
Loss
10 60 0.15 9.00
11 40 0.20 8.00
12 20 0.40 8.00
13 0 0.25 0.00
Total Loss 25.00

The Binomial Distributions
• Widely used probability distribution of a discrete random variable
• Describes discrete, non continuous, data, resulting from an
experiment also non as Bernoulli process
Use of Bernoulli Process (Flipping a coin)
• Each trail has fixed 2 outcomes (Heads or Tails)
• Probability of the outcome remains fixed at any given time (0.5)
• All the trails are statistically independent

Binomial or Bernoulli’s Formula
• If p is the probability of success
• If q is the probability of failure = 1-p
• Then probability of r successes in n trails =
𝑛!
𝑟! 𝑛 − 𝑟 !
𝑝 𝑟
𝑞 𝑛−𝑟

What is the probability of getting 2 success in 3
experiments?
• Probability of 2 successes in 3 experiments
• p = 0.5; q = 1-p = 0.5
• r = 2; n = 3
• Probability =
3!
2! 3−2 !
0.52
0.53−2
= 0.375

But how to create a distribution?

How probability effects the distribution?

How count of experiments effect the distribution?

Measures of central tendency and Dispersion
• Mean of a Binomial Distribution = µ = np
• Standard Deviation of a binomial Distribution = σ = 𝑛𝑝𝑞

Problem 1
Calculate the binomial distribution with n=7 and p=0.2
a) P(r=5)
b) P(r>2)
c) P(r<8)
d) P(r>=4)

Problem 1
Calculate the binomial distribution with n=7 and p=0.2
a) P(r=5)
a) 0.0043
b) P(r>2)
a) 0.1480
c) P(r<8)
a) 1
d) P(r>=4)
a) 0.0333

The Poisson Distribution
• Multiple outcomes
• The mean number of arrivals per unit time be estimated from past
data
• The probability that exactly 1 arrival will occur in one interval a very
small number and constant
• The probability that 2 or more arrivals will occur in 1 interval such a
small number that we can assign it a 0 value
• The number of arrivals per interval independent of time
• The number of arrivals per interval not dependent on the number of
arrivals in any other interval

Poisson’s Formula
• Probability of a discrete random variable
• Lambda = the mean number of occurrences in a interval
• e = Natural logarithmic constant = 2.718
• P(x) =
𝐿𝑎𝑚𝑏𝑑𝑎 𝑥 ∗ 𝑒−𝑙𝑎𝑚𝑏𝑑𝑎
𝑥!

The Poisson Distribution Graph

The Normal Distribution
• Used for Continuous random variable
• The curve has a single peak, a bell shaped curve
• Mean lies at the centre of the normal curve
• Mean, Median and mode are the same value
• The tails of the curve will never touch the horizontal axis

Normal Curve

Areas under Normal Curve

Normal Random Variable
• x = value of a random variable
• µ = Mean of the distribution of this random variable
• σ = Standard Deviation of distribution
• z = No.of standard deviations from the x to mean of the distribution
z =
𝑥−µ
σ

Probability Calculation
At a Driving tests transporting good is task that needs to be deliver by
500 Hrs of time but Jim took 650 Hrs to complete the transport. What
is the probability of Jim passing the test?
z =
𝑥−µ
σ
z =
650−500
100
z= 1.5 Standard Deviations

Probability Check from Standard Deviations
For 1.50 Standard
Deviations, the probability
of Jim passing the test is
little over 0.4 (~0.4332)

Probability Distributions Guide for Decision Making

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