EC255 Week2b

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Chapter 4
Probability
EC255: Managerial Statistics
Fall 2018
Tiffany Bayley
Learning Objectives
•Understand the different ways of assigning probability
•Understand and apply marginal, union, joint, and conditional probabilities
•Select the appropriate law of probability to solve problems
•Solve problems using the laws of probability
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Probability Matrices
•Clean way to show marginal and intersection probabilities
•Easy way to compute union and conditional probabilities
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How to Create a Probability Matrix
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How to Create a Probability Matrix
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How to Create a Probability Matrix
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How to Create a Probability Matrix
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Example 1:
Let’s consider mutual fund managers and their education (MBA from a Top-20 school
or not) and how their funds perform (outperform the market or not). Let:
•1 denote if the fund manager has an MBA from a Top-20 school and
•2 denote if the fund manager does not have an MBA from a Top-20 school.
•1 denote if their mutual fund outperforms the market and
•2 denote if their mutual fund does not outperform the market.
You are given  1 ∩ 1 = 0.11,  2 ∩ 1 = 0.06,  1 ∩ 2 = 0.29 and  2 ∩
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Example 1: Solution
a)
• 1 denote if the fund manager has an MBA from
a Top-20 school and
The probability that a fund manager has
an MBA from a Top-20 school
P(A1) = 0.40
• 2 denote if the fund manager does not have an
MBA from a Top-20 school.
• 1 denote if their mutual fund outperforms the
market and
a)
• 2 denote if their mutual fund does not
outperform the market.
P(B2): 0.83
•  1 ∩ 1 = 0.11
•  2 ∩ 1 = 0.06
•  1 ∩ 2 = 0.29
•  2 ∩ 2 = 0.54
EC255: Chapter 4


Total

0.11
0.29
0.40

0.06
0.54
0.60
Total
0.17
0.83
1.00
The probability that a fund will
underperform the market.
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Using the Probability Matrix to Solve Problems…
Law of Addition:
•Calculate P(Red or Ace)
28/52
Color
Type
Red
Black
Total
Ace
2
2
4
Non-Ace
24
24
48
Total
26
26
52
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Using the Probability Matrix to Solve Problems…
Conditional Probability:
•Of the cars on a used car lot
• 70% have air conditioning (AC)
• 40% have bluetooth (BT)
• 20% of the cars have both
BT
Non-BT
Total
AC
0.20
0.50
0.70
Non-AC
0.20
0.10
0.30
Total
0.40
0.60
1.00
•What is the probability that a car has
bluetooth, given that it has AC ?
0.2/0.7
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Using the Probability Matrix to Solve Problems…
Conditional Probability:
•Out of a target audience of 2,000,000
• ad A reaches 500,000 viewers
• ad B reaches 300,000 viewers
• both ads reach 100,000 viewers
ad B
ad A
Non-ad B
Total
100,000
500,000
300,000
2,000,000
Non-ad A
Total
•What is the probability of reaching a person
with ad A, given that ad B has been seen
already?
=(100,000/2,000,000)/(300,000/2,000,000)
= 1/3
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Concept Check W2-4b #4
• A survey found 43% of Canadians expect to
save more money next year than they did
last year
• 45% plan to reduce debt next year
• Of those who expect to save money, 81%
plan to reduce debt next year
• A Canadian is selected at random
• What is the probability that this person
expects to save more money and plans to
reduce debt?
• P(M) = 0.43
• P(R) = 0.45
• P(R|M) = 0.81
• P(M n R) = P(R|M) x P(M)
= 0.81 x 0.43
= 0.3483
a) 0.3483
b) 0.88
c) 0.4345
d) 0.19
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Concept Check W2-4b #5
• A survey found 43% of Canadians expect to
save more money next year than they did
last year
• 45% plan to reduce debt next year
• Of those who expect to save money, 81%
plan to reduce debt next year
• A Canadian is selected at random
• What is the probability that this person
expects to save more money or plans to
reduce debt?
a) 0.3483
b) 0.5317
c) 0.5755
d) 0.45
EC255: Chapter 4
• P(M) = 0.43
• P(R)=0.45
• P(R|M) = 0.81
P(M U R) =
P(M) + P(R) - P(MnR)
= 0.43 + 0.45 – 0.3483
=0.5317
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• A survey found 43% of Canadians expect to
save more money next year than they did
last year
• 45% plan to reduce debt next year
• Of those who expect to save money, 81%
plan to reduce debt next year
• A Canadian is selected at random
• What is the probability that this person
neither expects to save more money nor
plans to reduce debt?
a) 0.3483
b) 0.537
c) 0.5755
d) 0.4683
EC255: Chapter 4
• P(M) = 0.43
• P(R)=0.45
• P(R|M) = 0.81
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P(M’ n R’)
= 1– P(M U R)
= 1 – 0.5317
= 0.4683
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