Basics of Probability

Trial or Experiment

• Experiment - a process that results in a particular outcome or “event”.

• Simple event (or sample point), Ei – an event that can’t be decomposed into multiple individual outcomes.

• Sample space, S - The set of all possible sample points for the experiment.

• Event, Ai - a subset of the sample space.

Likelihood of an Outcome

• define the "likelihood" of a particular outcome or “event”

where an event is simply a subset of the sample space.

• Assuming each sample point is equally likely,

number of elements in event ( )

number of elements in sample space

AP A

1 1( )

number of elements in sample spaceiP ES

A Simple Experiment

• jar contains 3 quarters, 2 dimes, 1 nickels, and 4 pennies,

• consider randomly drawing one coin.The sample space:

• Let A be the event that a quarter is selected

1 2 3{q , q , q }A

Drawing a Quarter?

• Randomly draw a coin from the jar...

• There are 3 quarters among the 10 coins:

• Assuming each coin is equally likely to be drawn.

3( ) 0.3

10

AP A

S

Roll the Dice

• Using the elements of the sample space: (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

• Considering the sum of the values rolled,

Roll the Dice

• Using the elements of the sample space:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

• Count the members for this event.

Roll the Dice

• Using the elements of the sample space: (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Likelihood one of the faces shows a “2” ?

P( a “2” is rolled ) =

2). Summing over all the sample points in the sample space…

“a 100% chance one of the outcomes occurs” and…

Properties of a Probability

• Each outcome Ai in the sample space is assigned a probability value P( Ai )

1). “between a 0% and a 100% chance of occurring”: 0 ( ) 1iP A

( ) 1i

iE S

P E

Properties of a Probability

…and 3). When a set of events A1, A2, A3,… is

pairwise mutually exclusive…

1 2 3( ) ( )ii

P A A A P A

P( “2” is rolled OR sum is greater than 8 ) =

The Sample-Point Method

• Define the sample space: describe and list the simple events, being careful not to include any compound events.

• Assign a probability to each sample point, satisfying the “properties of a probability”.

• Define the event of interest, A, as a set of sample points.

• Compute P(A) by summing the probabilities of sample points in A.

Sticky Spinner

• Suppose a game uses a spinner to determine the number of places you may move your playing piece.

• Suppose the spinner tends to stop on “3” and “6” twice as often as it stops on the other numbers.

• What is the probability of moving a total of 9 spaces on your next 2 spins?

1 2

3

45

6

Multiplication Principle

( called “mn rule” in text )

Cross Product and Power Set

• By the multiplication principle,if | A| = m and | B| = n, then | A x B| = mn.

• By the multiplication principle, if | A| = n, then ( ) 2 2 2 2 2n

n times

A

“Decision Tree”

Total of 24

different systems

Computer 3 choices

Scanner 4 choices

Printer2 choices

Addition Principle

For any two sets A and B,

| | A B A B A B

In particular, if A and B are disjoint sets, then

| | A B A B

Extended to 3 sets…

| |

A B C A B C

A B A C B C

A B C

May generalize further for any n sets.

So for probability…

• Leads to an “addition rule for probability”:

( )

| | | | | |

| |

| | | | | |

| | | | | |

( ) ( ) ( )

A BP A B

S

A B A B

S

A B A B

S S S

P A P B P A B

Additive Rule of Probability

and if events A and B are mutually exclusive events, this simplifies to

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

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

Either Way

• Note we can do addition first, then convert to a probability ratio:

• Or we can construct the probabilities,then do addition:

| | | | | |( )

| |

A B A BP A B

S

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

Compute the Probability

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

( 6) ?P doubles sum

Compute the Probability

• Given the following probability table:

sedan mini-van truck totals

male .16 .10 .20 .46

female .24 .22 .08 .54 .40 .32 .28 1.00

• If one of the owners is randomly selected…

(female sedan owner)P

Counting Permutations

The number of ways to choose and arrange any r objects chosen from a set of n available objects, when repetitions are not allowed.

“Permutations of n objects, taken r at a time”

!

( )!n

r

nP

n r

Gold, Silver, Bronze

• Consider the top 3 winners in a race with 8 contestants. How many results are possible?

Or equivalently,

83

8! 8!(8)(7)(6) 336

(8 3)! 5!P

Calculate it

• Calculators have a built-in feature for these computations (labeled as nPr ).

• Use the MATH button and PRB submenu.

• To compute the value

we simply enter: 8 nPr 3

83 336P

Compare the 2 cases

• Case 1:• If a president, VP, and

treasurer are elected, how many outcomes are possible?

• (select and arrange 3, order is important)

• 16 x 15 x 14 = 3360 pres. VP treas.

Consider a club with 16 members:• Case 2:• If a group of 3 members

is chosen, how many groups are possible ?

• (a choice of 3 members, order is not important)

• Since we don't count the different arrangements, this total should be less.

Adjust the total

Case 1: Case 2:

Given one group of 3 members, such as Joe, Bob, and Sue, 6 arrangements are possible: ( Joe, Bob, Sue), ( Joe, Sue, Bob), ( Bob, Joe, Sue)

( Bob, Sue, Joe), ( Sue, Joe, Bob), ( Sue, Bob, Joe)

Each group gets counted 6 times for permutations.

Divide by 6 to “remove this redundancy”.

16 163 3

33603360 560

6P C

Counting Combinations

• “Combinations of n objects, taken r at a time” when repetitions are not allowed

Often read as “n, choose r"

!

! !( )!

nn rr

P nC

r r n r

Sometimes denoted as ,, or n r

nC

r

All Spades?

• For example, in a 5-card hand, P( all 5-cards drawn are spades)

4 Spades, and a Non-Spade?

• For example, in a 5-card hand, P(exactly 4 spades in 5-card hand)?

(hands with 4 spades, 1 non-spade)

( possible 5-card hands)

n

n

13,4 39,1

52,5

(715)(39)0.01072929

2598960

C C

C

All Possible Cases?

Consider the possible number of spades:• P(all 5 spades) = 0.00049520• P(exactly 4 spades) = 0.01072929• P(exactly 3 spades) = 0.08154262• P(exactly 2 spades) = 0.27427971• P(exactly 1 spade) = 0.41141957• P(no spades) = 0.22153361

1.000

Exactly 3 Face Cards?

• “3 face cards” implies other 2 cards are not face cards

• P( 5-card hand with exactly 3 face cards) = ?

Probable Committee?

• If a 3-person committee is selected at random from a group of 6 juniors and 9 seniors, what is the probability that exactly 2 seniors are selected?

• Setup the ratio, this type of committee as compared to all possible 3-person committees.

Binomial Coefficients

• Recall the Binomial Theorem: For every non-negative integer n…

0

( )n

n n n k kk

k

x y C x y

Remember “Pascal’s Triangle”?

Multinomial Coefficients

• “Ways to partition n objects into k groups” when repetitions are not allowed

Here the groups are of size n1, n2, …, and nk

such that n1 + n2 + … + nk = n.

1 2 1 2

!

! ! !k k

n nn n n n n n

Expanding a Multinomial

31 2

1 2 3 1 2 3

( ) nn nn

n n n n

nx y z x y z

n n n

• Using the multinomial coefficients

8( 3 )x y z Determine the coefficient of the x3y3z2 term.

In the expansion of the multinomial