COM2023
Mathematics Methods for Computing II
Lecture 5& 6
Gianne Derks
Department of Mathematics (36AA04)
http://www.maths.surrey.ac.uk/Modules/COM2023
Autumn 2010
Use channel 04 on your EVS handset
Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Probability
Some definitions . . . . . . . . . . . . . .
More definitions . . . . . . . . . . . . . .
Example: Throwing an unbiased die .
Throwing a fair dodecahedron die . .
Rules
Rules of Probability . . . . . . . .
Exercise on rules of probability
Partition Law . . . . . . . . . . . .
Exercise on Partition Law . . . .
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Overview
Probability
●
Introduction: notation, definitions and an example
●
Recall definitions and example
●
Rules of probability and Venn diagrams
●
Conditional probability
Introduction to probability (§3.1)
Some definitions
●
The sample space is the set of all possible outcomes.
For example the sample space related to observations of a traffic light is
S = {red, orange, green, red-orange}.
●
An event is a collection of some outcomes in the sample space.
For example, for the sample space above, a possible event would be the event
A = {orange, red-orange}.
●
The complement of the event A are all outcomes in S, that
are not in A. The complement of A is denoted by Ac .
For example, with S and A as defined above, Ac =
{red, green}.
A
A
2
c
More definitions
●
●
The union union of two events A and B is denoted by A ∪ B. It
contains all outcomes in A or in B or in both.
For example, if B = {red, orange}, then A ∪ B =
{red, orange, red-orange}.
The intersection of two events A and B is denoted by A ∩ B.
It contains all outcomes which are both in A and in B.
In the example, A ∩ B = {orange}.
A
B
A
B
●
The events A and B are said to be disjoint if they have no common outcome, i.e., A ∩ B is
empty.
●
The probability that the event A happens is denoted by P (A).
Example: Throwing an unbiased die
Sample space: S = {1, 2, 3, 4, 5, 6}. First some events:
• Event A = {2, 4, 6}
⇒ P (A) =
• Event B = {5, 6}
⇒ P (B) =
1
2
1
3
Related events:
1
2
2
3
• Event Ac = {1, 3, 5}
⇒ P (Ac ) =
• Event B c = {1, 2, 3, 4}
⇒ P (B c ) =
• Event A ∪ B = {2, 4, 5, 6}
⇒ P (A ∪ B) =
• Event (A ∪ B)c = {1, 3}
⇒ P ((A ∪ B)c ) =
• Event A ∩ B = {6}
⇒ P (A ∩ B) =
• Event (A ∩ B)c = {1, 2, 3, 4, 5}
⇒ P ((A ∩ B)c ) =
3
2
3
1
3
1
6
5
6
Throwing a fair dodecahedron die
Sample space: S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Some events:
• Event A = {1, 3, 5, 7, 9, 11}
⇒ P (A) =
• Event B = {2, 3, 5, 7, 11}
⇒ P (B) =
1
2
5
12
Related events:
1
2
7
12
• Event Ac = {2, 4, 6, 8, 10, 12}
⇒ P (Ac ) =
• Event B c = {1, 4, 6, 8, 9, 10, 12}
⇒ P (B c ) =
• Event A ∪ B = {1, 2, 3, 5, 7, 9, 11}
⇒ P (A ∪ B) =
• Event (A ∪ B)c = {4, 6, 8, 10, 12}
⇒ P ((A ∪ B)c ) =
• Event A ∩ B = {3, 5, 7, 11}
⇒ P (A ∩ B) =
• Event (A ∩ B)c = {1, 2, 4, 6, 8, 9, 10, 12}
⇒ P ((A ∩ B)c ) =
7
12
5
12
1
3
2
3
Rules of Probability (§3.2)
Rules of Probability
General rules
●
P (S) = 1, where S is the sample space;
●
P (∅) = 0, where ∅ is an empty event;
●
0 ≤ P (A) ≤ 1, for any event A;
●
P (Ac ) = 1 − P (A), for any event A;
Addition law
●
P (A ∪ B) = P (A) + P (B), whenever A ∩ B = ∅;
●
P (A ∪ B) = P (A) + P (B) − P (A ∩ B).
A
B
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Exercise on rules of probability
Manufactured components can have two types of common defect, call them defect I and defect II.
Sampling has shown that
●
5% of the components have defect I;
●
2% of the components have defect II;
●
0.5% of the components have both defects.
A component is selected randomly. What is the probability that this component does not have defect
I nor defect II?
Answer: P ((A ∪ B)c ) = 1 − [P (A) + P (B) − P (A ∩ B)] = 93.5%
Partition Law
●
Partition the sample space S into two events B and B c :
P (A) = P (A ∩ B) + P (A ∩ B c ).
●
Partition the sample space S into three events B1 , B2 , and B3 :
P (A) = P (A ∩ B1 ) + P (A ∩ B2 ) + P (A ∩ B3 ).
●
Partition sample space S into k events B1 , B2 , . . . Bk :
P (A) =
k
X
i=1
5
P (A ∩ Bi ).
Exercise on Partition Law
Light bulbs are produced in four colours: blue, green, red and orange and equal amounts of each
colour are produced. Sampling has shown that
0.25*98%
●
2% of the blue bulbs is faulty;
●
5% of the green bulbs is faulty;
●
7% of the red bulbs is faulty;
Blue
0.25*95%
Green
0.25*2%
0.25*5%
0.25*7%
0.25*1%
Red
Orange
1% of the orange bulbs is faulty.
0.25*93%
0.25*99%
Make a Venn diagram with all colours and indicate the proportion of faulty ones in each colour.
●
●
If you get an arbitrary light bulb, what is the change of getting a faulty blue one? 2% × 0.25
= 0.5%
●
How about the other colours? gr: 1.25%, red: 1.75%, or: 0.25%
●
Take an arbitrary light bulb; what is the change of getting a faulty one? P(faulty)= 0.5% +
1.25% + 1.75% + 0.25% = 3.75%
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