Chapter 11 Probability(Week 12-13)

Lecture 2: - Probability of Events

- Counting Rule

Apr 8, 2023 1

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Learning Objectives

At the end of the session student should be able to:

- Determine the probability of events- Interpret and understand the properties of probability

and utilize them in solving problems- Use multiplication rule (one type of counting rule) to find

the number of ways and possible outcomes of a given

random experiment.

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Probability of Events

The objective of probability is to assign to each event, say E, a number P(E), called the probability of E which will give a precise measure of the chance that E will occur.

The probability of any event E, P(E) = n/N

where n : number of times we observe the event (frequency)

N : a very large number of trials

Q: In the experiment tossing a die repeately, in the long run, what would we expect P(E1) (probability of number 1 occurs) equal to?

( A: 1/6)

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Definition of Probability

A probability P is a rule ( or function) which assigns a number between 0 and 1 to each event and satisfies:

0 ≤ P(E) ≤ 1 for any event E

P(Ø ) = 0 , P(S) = 1,

If A1 , A2 , … is an infinite collection of mutually exclusive events, then

1 2 1 2( ...) ( ) ( ) ...P A A P A P A

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The probability of the complement of any event A is given as

Example: If P(rain tomorrow) = 0.6 then P(no rain tomorrow) = 0.4

Other notations for complement: Ac or Ā

( ') 1 ( )P A P A

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General Addition Law

Let A and B be two events defined in a sample space S.

If two events A and B are mutually exclusive, then

Thus

This can be expanded to consider more than two mutually exclusive events.

P(A B) P(A) P(B) P(A B)

P(A B) 0

P(A B) P(A) P(B)

Apr 8, 2023

Example 3: Samples of building materials from three suppliers are

classified for conformance to air-quality specifications.

The results from 100 samples are summarized as follows:

 

Let A denote the event that a sample is from supplier R, and B denote

the event that a sample conforms to the specifications. If sample is

selected at random, determine the following probabilities: (a) P(A) (b) P(B) (c) P(B’) (d) P(AUB) (e) P(AB) (f) P(AUB’)

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Conforms

Yes No

SupplierR 30 10

S 22 8

T 25 5

Apr 8, 2023

A : sample from supplier R, and

B : sample conforms to the specifications

a. P(A) = 40/100 = 0.4

b. P(B) = 77/100 = 0.77

c. P(B’) = 23/100 = 0.23

d. P(AUB) = 87/100 = 0.87

e. P(AB) = 30/100 = 0.3

f. P(AUB’) = 53/100 = 0.53

8Apr 8, 2023

Example 4.

In a residential suburb, 60% of all households subscribe

to the metro newspaper published in a nearby city, 80%

subscribe to the local paper, and 50% of all households

subscribe to both papers. Draw a Venn diagram for this

problem. If a household is selected at random, what is

the probability that it subscribes to:

a) at least one of the two newspapers b) exactly one of the two newspapers

 9Apr 8, 2023

A : subscribe metro

B : subscribe local

10Apr 8, 2023

A

B

0.1

0.3

0.1

0.5

a. P(AUB) = 0.9

b. P(AB’UA’UB)=P(AB’)+P(A’B = 0.1+0.3 = 0.4

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Counting Rule

P(A) = n(A) / n(S)

The probability of an event A equals the number of outcomes (sample points) contained in A divided by the total number of possible

outcomes. Important condition: all outcomes are equally likely to occur. Inefficient when n(S) is large.

Counting Rule: Eliminates the need for listing each simple event and help to easily assign probabilities to various events when the outcomes are equally likely.

Especially helpful if the sample space is quite large.

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Counting Rule: Product (Multiplication) Rule If there are k elements ( or things) to choose

and there are n1 choices for the first element,

n2 for the second element, and so on to nk choices for the kth element,

then the number of possible ways of selecting them is

Only applies when elements are different or the order of elements matters.

1 2 ... kn n n

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Example 5: In a student organization election, we want to

elect one president from five candidates, one

vice president from six candidates, and one

secretary from three candidates. How many

possible outcomes?

13Apr 8, 2023

90 5X6X3 are outcomes Possible

:Solution

Example 6:A chemical engineer wishes to conduct an

experiment to determine how these four factors

affect the quality of the coating. She is interested

in comparing three charge levels, five density

levels, four temperature levels, and three speed

levels. How many experimental conditions are

possible?

14Apr 8, 2023

180 3X5X4X3 are outcomes Possible

:Solution