Discrete Probability

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CountingSummation Principle

Multiplication Principle

Permutation

Combination

Pigeon Hole Principle

Probability

6.0 Discrete Probability

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Everything you have learned about counting constitutes the basis for computing the probability of events to happen.

Used to assign values to the experiment/test that does not guarantee to return the same value on repeated test

.

Different from predicted test that always gives the same result.

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Probability : Introduction

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Random Test Test or process that give

unpredictable result

Sample possible result from certain test

Sample Space (from random test) a set that consist

of all samples from certain test

Event A statement that explain the result of a test;

the result is either true or false

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Probability

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Example:

Random Test: flip a coin twice and record head-

tail sequence.

Samples: HH, TT, TH, HT

Sample Space: { HH, TT, TH, HT }

(all 4 possibilities)

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Probability: Sample Space

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Test Example: Toss 20 cent and 50 cent coins.

1. If the test is to count number of heads, then the sample space is { 0,1,2 }

2. If the test is to record head-tail sequence of 20 cent followed by 50 cent, therefore the sample space is { HH, HT, TH, TT }

3. If the test is to record whether the result is the same pair or not (e.g. HH or TT is consider same pair), therefore the sample space is {Same, Different}

Sample space depends on what need to be observed

(related to the event) from the test.

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Probability: Event

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Example:

Random Test: Flip a coin twice and record

head-tail sequence.

Sample space: S = { HH, HT, TH, TT }

Event A: get at least one head

A = { HH, HT, TH }

Event B: did not get tail

B = { HH }March 8, 2015

Probability: Event

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Sample space: S = { HH,HT,TH,TT }

Event A A = { HH, HT, TH }

Event B B = { HH }

Since the sample space and event are set, therefore

the set operation (intersection, union, etc.) can be applied

on sample spaces and events.

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Probability: Event

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Event A = {HH, TT} (same pair)

Event C = {HT, TH} (only one head)

Based on the above two events we can

generate new events:

Event A and C (same pair and only one head)

Event A or C (same pair or only one head)

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Probability: Event

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Event A and C = { }

Event A or C = { }

Using set operation:

A or C = A C = { HH,HT,TH,TT }

A and C = A C = { }

Union: X Y X or Y

Intersection: X Y X and Y

Complement: X not X

Event that did not intersect is known as mutually exclusive :

X Y =

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Probability: Event

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Event is always a subset to a sample space

Empty Set () and sample space set is also an event, according to the definition.

If S is a sample space, therefore S is a definite event and an empty set is an impossible event

Note: for sample space of size n, there are 2n events (remember power set? And counting technique on how to count it?)

sample space: S = { HH, HT, TH, TT }

event X X = { }

event Y Y = { HH, HT, TH, TT }

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Probability

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Example: The test is to flip a coin once and observe the result.

Get the probability head, H, happen.

Assume to obtain H and T are equally likely (that means we have a fair coin)

In word: from 2 possible events (the whole sample space) there is 1 probability that event H can happen

In probability: p(H) = 1 choicefrom 2

=1

2March 8, 2015

Probability

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Note:

Equal likely event is the assumption that each

case of the sample space is equally likely to

happen.

Equal likelihood case might not necessarily occur.

Example, in the case of cheated coin

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Probability

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Each event E is given a value P(E),

known as event probability.

P(E) reflects the assumption made

about the likelihood of event E to

happen.

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Probability

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Determine the Probability

Sample space: { x1, x2, x3, . . . , xn }

Determine the probability:

p(x1)= p1, p(x2) = p2, . . . p(xn)=pn

Characteristic in determining the probability:

1. 0 pi 1, i = 1, n

2. p1 + p2 + + pn = 1

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Probability

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Probability event

Event

A = { y1, y

2, y

3, y

m}

p(A) = mi=1 pi

where yihas probability

pito happen

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Probability

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Equally likely event probability

Sample space with N probability

It is said that there is 1 probability from N total possibility = 1/N

Event probability of size m has m probability from N total possibility or m/N

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Probability

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Example:

Toss a dice once and observe the result

1. How many possibility are there?

2. What is the probability of each?

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Discrete Probability

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Example 1:

An urn contains four blue balls and five red

balls. What is the probability that a ball

chosen from the urn is blue?

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Solution:

1. There are nine possible outcomes, and

2. The event blue ball is chosen comprises four of these

outcomes.

3. Therefore, the probability of this event is 4/9 or

approximately 44.44%.

Note: S = { b1, b2, b3, b4, r1, r2, r3, r4, r5 }

Note: E = { b1, b2, b3, b4 }

Therefore p(E) = 4/9

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Example 2:

What is the probability of winning the lottery that

is, you have to pick the correct set of six numbers

out of 49 numbers?

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Solution:

1. There are C(49, 6) possible outcomes.

2. Only one of these outcomes will actually make you

win the lottery.

|S| = C(49, 6), |E| = 1

Therefore;

p(E) = 1/C(49, 6) = 1/13,983,816

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Complimentary Events

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Let E be an event in a sample space S. The probability of an event not E( E), the complimentary event of E, is given by

p(E) = 1 p(E).

This can easily be shown:

This rule is useful if it is easier to determine the probability of the complimentary event than the probability of the event itself.

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Example 1:

A sequence of 10 bits is randomly generated.

What is the probability that at least one of

these bits is zero?

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Solution:

1. There are 210 = 1024 possible outcomes of

generating such a sequence.

2. The event E, none of the bits is zero, includes only one of these outcomes, namely the sequence

1111111111.

|S|=1024, |E |=1

Therefore, p(E) = (1/1024).

Now p(E) can easily be computed as

p(E) = 1 p(E) = 1 (1/1024) = 1023/1024.March 8, 2015


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Let E1 and E2 be events in the sample space S. Then we have:

p(E1 E2) = p(E1) + p(E2) - p(E1 E2)

Does this remind you of something?

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Example:

What is the probability of a positive integer

selected at random from the set of positive

integers not exceeding 100 to be divisible by 2

or 5?

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Solution:

1. E2: integer is divisible by 2

2. E5: integer is divisible by 5

E2 = {2, 4, 6, , 100}

|E2| = 50, |S|=100

p(E2) = 0.5

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E5 = {5, 10, 15, , 100}

|E5| = 20

p(E5) = 0.2

E2 E5 = {10, 20, 30, , 100}

|E2 E5| = 10

p(E2 E5) = 0.1

p(E2 E5) = p(E2) + p(E5) p(E2 E5 )

=p(E2 E5)

= 0.5 + 0.2 0.1

= 0.6March 8, 2015


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What happens if the outcomes of an experiment are not equally likely?

In that case, we assign a probability p(s) to each outcome s S, where S is the sample space.

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Two conditions have to be met:

(1): 0 p(s) 1 for each s S, and

(2): sS p(s) = 1

Meaning,

each probability must be a value between 0 and 1, and

the probabilities must add up to 1, because one of the outcomes is guaranteed to occur.

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How can we obtain these probabilities p(s) ?

The probability p(s) assigned to an outcome s equals

the limit of the number of times s occurs divided by the

number of times the experiment is performed.

Once we know the probabilities p(s), we can compute

the probability of an event E as follows:

p(E) = sE p(s)

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Example 1:

A dice is biased so that the number 3 appears twice as often as each other number. What are the probabilities of all possible outcomes?

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Solution:

There are 6 possible outcomes s1, , s6.

p(s1) = p(s2) = p(s4) = p(s5) = p(s6)

p(s3) = 2p(s1)

Since the probabilities must add up to 1, we have:

5p(s1) + 2p(s1) = 1

7p(s1) = 1

p(s1) = p(s2) = p(s4) = p(s5) = p(s6) = 1/7,

p(s3) = 2/7March 8, 2015


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Example 2:

For the biased dice from Example 1, what is the

probability that an odd number appears when

we roll the dice?

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Solution:

Eodd = {s1, s3, s5}

Remember the formula

p(E) = sE p(s).

p(Eodd) = sEodd p(s) = p(s1) + p(s3) + p(s5)

p(Eodd) = 1/7 + 2/7 + 1/7 = 4/7 = 57.14%

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Conditional Probability

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If we toss a coin three times, what is the probability that an odd number of tails appears (event E), if the first toss is a tail (event F) ?

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If the first toss is a tail, the possible sequences are TTT, TTH, THT, and THH.

In two out of these four cases, there is an odd number of tails.

Therefore, the probability of E, under the condition that F occurs, is 0.5.

We call this conditional probability.

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If we want to compute the conditional probability of E

given F, we use F as the sample space.

For any outcome of E to occur under the condition that F

also occurs, this outcome must also be in E F.

Definition: Let E and F be events with p(F) > 0.

The conditional probability of E given F, denoted by

p(E | F), is defined as

p(E | F) = p(E F) / p(F)

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Example:

What is the probability of a random bit string of

length four to contain at least two consecutive 0s,

given that its first bit is a 0 ?

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Solution:1. E: bit string contains at least two consecutive 0s

2. F: first bit of the string is a 0

3. We know the formula p(E | F) = p(E F)/p(F).

E F = {0000, 0001, 0010, 0011, 0100}p(E F) = 5/16p(F) = 8/16 =

p(E | F) = (5/16)/(1/2) = 10/16

= 0.625

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Discrete Probability

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