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Understanding Probability Theory: Random Experiments, Outcomes, and Counting - Prof. David, Study notes of Mathematical Statistics

This chapter introduces the concept of a random experiment and its related terminology, such as elementary outcomes and sequences of outcomes. It also covers methods for counting the number of possible outcomes when there are several steps in an experiment, using examples of rolling dice and drawing cards. The document further discusses the importance of events and their probabilities in probability theory.

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2010/2011

Uploaded on 02/12/2011

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CHAPTER 4
The notion of a random experiment is critical before taking up the subject of probability.
Random experiment: A well-defined procedure with a well-defined outcome (generally with an
uncertain outcome).
Common examples: flip a coin, roll a single die, roll two dice, draw two a card at random from a
standard card deck
Challenge for probability: When an experiment is considered, how many possible outcomes are
There? An elementary outcome is one which cannot be broken down further. So, if a single die is
rolled, the elementary outcomes are the numbers: 1, 2, 3, 4, 5, 6.
Note: you might be interested in classifying outcomes broadly such as “roll a even number,”
“number is greater than 2,” or “number is less than 5.” But all these outcomes are not
elementary, because they are comprised of more than one elementary outcome. For example, the
outcome: “roll an even number” is actually made up of outcomes 2, 4, and 6.
Before we tackle probability, we will consider the case when experiments are made up of a
sequency of several steps.
Examples: Roll two dice (die A and die B) – even if the dice are rolled simultaneously, the overall
outcome is made up of a pair of numbers, such as (a,b) = (1, 3) where die A = 1 and die B=3.
Or, flip 3 coins. Or draw 2 cards at random from a deck.
Let’s talk about a general method for counting the number of possible outcomes when there are
several steps in an overall experiment. Important: the outcome is considered to be the entire
sequence of outcomes at all steps in the experiment.
Suppose we flip a coin twice. The possible sequences of outcomes can be illustrated by a chain:
HH, TT, HT, TH (notice that HT is a different sequence than TH)
Notice that the total number of possible chains is 2*2=4 (2 outcomes are possible at each step in
the chain)
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CHAPTER 4

The notion of a random experiment is critical before taking up the subject of probability.

Random experiment: A well-defined procedure with a well-defined outcome (generally with an

uncertain outcome).

Common examples: flip a coin, roll a single die, roll two dice, draw two a card at random from a

standard card deck

Challenge for probability: When an experiment is considered, how many possible outcomes are

There? An elementary outcome is one which cannot be broken down further. So, if a single die is

rolled, the elementary outcomes are the numbers: 1, 2, 3, 4, 5, 6.

Note: you might be interested in classifying outcomes broadly such as “roll a even number,”

“number is greater than 2,” or “number is less than 5.” But all these outcomes are not

elementary, because they are comprised of more than one elementary outcome. For example, the

outcome: “roll an even number” is actually made up of outcomes 2, 4, and 6.

Before we tackle probability, we will consider the case when experiments are made up of a

sequency of several steps.

Examples: Roll two dice (die A and die B) – even if the dice are rolled simultaneously, the overall

outcome is made up of a pair of numbers, such as (a,b) = (1, 3) where die A = 1 and die B=3.

Or, flip 3 coins. Or draw 2 cards at random from a deck.

Let’s talk about a general method for counting the number of possible outcomes when there are

several steps in an overall experiment. Important: the outcome is considered to be the entire

sequence of outcomes at all steps in the experiment.

Suppose we flip a coin twice. The possible sequences of outcomes can be illustrated by a chain:

HH, TT, HT, TH (notice that HT is a different sequence than TH)

Notice that the total number of possible chains is 2*2=4 (2 outcomes are possible at each step in

the chain)

If you roll two dice (each with 6 sides), the number of total sequences is:

6*6=36 (6 outcomes are possible with each die)

If you draw 2 cards from a deck with replacement, you get: 52*52=2704 outcomes

If you draw 2 cards WITHOUT replacement, the result is: 52*51=2652 possible outcomes

Of particular interest are selections from a group without replacement that fall into 2 cases

(permutations and combinations):

You start with a group (say 12 people) and wish to select 4.

If we simply select 4 for a committee and only the identities of the individuals defines the outcome,

This is a combination:

The number of possible committees is:^495

4 C 12   

But, if we were to not only select 4 people, but ALSO indicate that Bill is President, Mary is Vice-

President, Julie is Treasurer, and Bob is Secretary, this is a Permutation. The answer is:

4 P 12   

For the rest of our discussion, let’s keep this two-dice experiment in mind as our example.

Probability of event A. The probability of an event generally assumes that all elementary events have

the same chance of occurring. This is labelled: P(A).

To compute P(A) is a simple matter of counting. Thinking of A as a set:

P(A) = nS)

n A ( ( )

Here, n(A) = the number of elementary outcomes in event A

n( S ) = number of elementary outcomes in S.

Given that we can think of events as sets, set operations (remember those?) can be useful. Set

operations include: Union A^ ^ B Intersection A  B Complement A '

If A = sum is 7 B = largest number is a 4

A  B = the event that either A OR B occurs

Then A^ ^ B = event that either the sum is 7 or the largest number is a 4

A  B = the event that both A AND B occur

A  B = the event that the sum is 7 and the largest number rolled is a r

A’ = the event that A does NOT occur

A’ = the event that the sum is not 7

The probabilities of these “compound” events can be determined by examining the sample space and

counting the number of outcomes where either A or B occurs (for A^ ^ B ), where A and B both occur

(for A  B ), and where A does not occur for A’.

Note: Mutually exclusive events are two events, A and B, such that P(^ A  B) ^0

For example: If you roll two dice and A = both number are even, B = sum of two dice is 9. Note, it

is impossible, than when you roll two dice, that both A and B will occur on the SAME experiment.

Or: P(^ A ^ B) ^0. So, there two particular events, A and B, are mutually exclusive.

CONDITIONAL PROBABILITY

Conditional probability is probability with a little extra information. If we know that some

event has occurred in an experiment, how does that impact our calculation of the likelihood

(probability) that some other event occurred on the same experiment?

Example:

The table belows records and summarizes data collected for a population of people, in which

the gender and eye color of each individual was recorded.

Gender Eye Color Female (^) Male

Blue 15

Brown 12

Other 28

Suppose, we define: A: Female B: Blue Eyes

Note, the total number of people in the population is 100. Suppose one person from this

population is selected at random.

What is P(A)? Add up the numbers in the Female column: 55

Then: P(A)=55/100 =.

What is P(B)? Add up the numbers in the Blue row: 30

So: P(B) = 30/100 =.

What is P(both numbers are even given that the sum of the two number is 8)?

Or, in symbols, P(B | A) (using the definition):

PA

PB A

P B A

Conditional probability can be useful in computing the probability of a compound experiment

(several steps) when an outcome at a step depends on previous steps.

Example:

Draw 2 cards at random without replacement.

What is the probability both cards are hearts?

If A = first card is a heart B = second card is a heart

We want to compute: P (^^ A  B )

By shifting terms in the definition of conditional probability, you can write:

P ( A  B ) P ( A )* P ( B | A )  

So, anytime you want to compute a probability of a chain of events (and probabilities at stages

in the process depend of previous steps), simply multiply the probabilities in the chain

(assuming that what must take place in the chain does so)

Independence

Interpretation: The fact that event B has occurred does not impact the probability of event A.

So: P^ ( B | A ) P^ ( B ) orequivalently, P(A  B)  P(A)*P(B)

Note: This is NOT the same idea as mutually exclusive events ( P^ (^ A  B )^0

REMEMBER THESE FORMULAE:

Some special probability rules that are helpful are:

P(A’) = 1 - P(A)

P ( A  B ) P ( A ) P ( B ) P ( A  B )