Probability Theory: Uncertain Quantities and Events - Prof. Johan Dorp, Assignments of Systems Engineering

Solutions to problems related to probability theory, focusing on uncertain quantities and events. It covers concepts such as joint and conditional probabilities, independent events, and the calculation of expected values and covariance.

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Pre 2010

Uploaded on 08/19/2009

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Solutions
7.1. We often have to make decisions in the face of uncertainty. Probability is a formal way to cope with
and model that uncertainty.
7.2. An uncertain quantity or random variable is an event that is uncertain and has a quantitative outcome
(time, age, $, temperature, weight, . . . ). Often a non-quantitative event can be the basis for defining an
uncertain quantity; specific non-quantitative outcomes (colors, names, categories) correspond to
quantitative outcomes of the uncertain quantity (light wavelength, number of letters, classification number).
Uncertain quantities are important in decision analysis because they permit us to build models that may be
subjected to quantitative analysis.
7.3. P(A and B) = 0.12 P(B
) = 0.35
P(A and B
) = 0.29 P(B | A) = 0.12
0.41 = 0.293
P(A) = 0.41 P(A | B) = 0.12
0.65 = 0.185
P(B) = 0.65 P(A
| B
) = 0.06
0.35 = 0.171
7.4. P(A or B) = P(A and B) + P(A and B
) + P(A
and B)
= 0.12 + 0.53 + 0.29 = 0.94
or P(A or B) = P(A) + P(B) - P(A and B)
= 0.41 + 0.65 - 0.12 = 0.94
or P(A or B) = 1 - P(A
and B
) = 1 - 0.06 = 0.94
7.5.
A
B
A and B
A and B A and B
From the diagram, it is clear that
P(A) = P(A and B) + P(A and B
)
and
P(B) = P(A and B) + P(A
and B).
But P(A or B) clearly equals P(A and B) + P(A and B
) + P(A
and B) because of property 2. Thus,
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Solutions

7.1. We often have to make decisions in the face of uncertainty. Probability is a formal way to cope with

and model that uncertainty.

7.2. An uncertain quantity or random variable is an event that is uncertain and has a quantitative outcome

(time, age, $, temperature, weight,... ). Often a non-quantitative event can be the basis for defining an

uncertain quantity; specific non-quantitative outcomes (colors, names, categories) correspond to

quantitative outcomes of the uncertain quantity (light wavelength, number of letters, classification number).

Uncertain quantities are important in decision analysis because they permit us to build models that may be

subjected to quantitative analysis.

7.3. P(A and B) = 0.12 P( B

P(A and B

) = 0.29 P(B | A) =

P(A) = 0.41 P(A | B) =

P(B) = 0.65 P(A

| B

7.4. P(A or B) = P(A and B) + P(A and B

) + P(A

and B)

= 0.12 + 0.53 + 0.29 = 0.

or P(A or B) = P(A) + P(B) - P(A and B)

= 0.41 + 0.65 - 0.12 = 0.

or P(A or B) = 1 - P(A

and B

A

B

A and B A and B A and B

From the diagram, it is clear that

P(A) = P(A and B) + P(A and B

and

P(B) = P(A and B) + P(A

and B).

But P(A or B) clearly equals P(A and B) + P(A and B

) + P(A

and B) because of property 2. Thus,

P(A or B) = P(A and B) + P(A and B

) + P(A

and B)

= P(A) + P(B) - P(A and B).

7.6.a. Joint. P(left-handed and red-haired) = 0.

b. Conditional P(red-haired | left-handed) = 0.

c. Conditional P(Cubs win | Orioles lose) = 0.

d. Conditional P(Disease | positive) = 0.

e. Joint P(success and no cancer) = 0.

f. Conditional P(cancer | success)

g. Conditional P(food prices up | drought)

h. Conditional P(bankrupt | lose crop) = 0.

i. Conditional, but with a joint condition: P(lose crop | temperature high and no rain)

j. Conditional P(arrest | trading on insider information)

k. Joint P(trade on insider information and get caught)

7.7. For Product B, EMV = $8M(0.38) + $4M(0.12) + 0(0.50) = $3.52M

Var(B) = 0.38(8 - 3.52)

= 13.8496 “Millions-of-dollars squared”

Standard Deviation for B = σ B =

13.8496 = $3.72M.

For Product C, there is no variation. Thus, Var(C) = 0 and σC = 0.

A A

B

B

P(A) = 0.42 is given, so P(A

) = 1 - P(A) = 1 - 0.42 = 0.

P( B

| A) = 1- P(B | A) = 1 - 0.66 = 0.

P( B

| A

) = 1- P(B | A

P(B) = P(B | A) P(A) + P(B | A

) P(A

P( B

) = 1 - P(B) = 1- 0.4222 = 0.

P(A | B) =

P(A and B)

P(B)

P(B|A )P(A)

P(B)

P(A

| B) = 1 - P(A | B) = 1 - 0.6566 = 0.

P(A | B

P(A and B

P( B

P( B

|A)P(A)

P( B