Download Basics of Probability

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”
number of elements in event A
P( A) 
number of elements in sample space
where an event is simply a subset of the
sample space.
• Assuming each sample point is equally likely,
1
1
P( Ei ) 

number of elements in sample space S
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 A  {q , q , q }
1
2
3
Drawing a Quarter?
• Randomly draw a coin from the jar...
• There are 3 quarters among the 10 coins:
A 3
P( A) 
  0.3
S 10
• Assuming each coin is equally likely to be
drawn.
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 ) =
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  P( A )  1
i
2). Summing over all the sample points in the
sample space…
 P( E )  1
Ei S
i
“a 100% chance one of the outcomes occurs”
and…
Properties of a Probability
…and
3). When a set of events A1, A2, A3,… is
pairwise mutually exclusive…
P( A1  A2  A3 
)   P( Ai )
i
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
2
1
3
6
5
4
• 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?
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,
n
then
( A)  2  2  2
n times
22
“Decision Tree”
Total of
24
different
systems
Computer
3 choices
Scanner
4 choices
Printer
2 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 B
P( 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
P( A  B)  P( A)  P( B)  P( A  B)
and if events A and B are mutually exclusive
events, this simplifies to
P( A  B)  P( A)  P( B)
Either Way
• Note we can do addition first,
then convert to a probability ratio:
| A|  | B |  | A B |
P( A  B) 
|S|
• Or we can construct the probabilities,
then do addition:
P( A  B)  P( A)  P( B)  P( A  B)
Compute the Probability
(1,1)
(2,1)
(3,1)
(4,1)
(5,1)
(6,1)
(1,2)
(2,2)
(3,2)
(4,2)
(5,2)
(6,2)
(1,3)
(2,3)
(3,3)
(4,3)
(5,3)
(6,3)
(1,4)
(2,4)
(3,4)
(4,4)
(5,4)
(6,4)
(1,5)
(2,5)
(3,5)
(4,5)
(5,5)
(6,5)
(1,6)
(2,6)
(3,6)
(4,6)
(5,6)
(6,6)
P(doubles  sum  6)  ?
Compute the Probability
• Given the following probability table:
sedan
male
female
.16
.24
.40
mini-van
truck
totals
.10
.22
.32
.20
.08
.28
.46
.54
1.00
• If one of the owners is randomly selected…
P(female  sedan owner) 
Counting Permutations
“Permutations of n objects, taken r at a time”
n!
P 
(n  r )!
n
r
The number of ways to choose and arrange
any r objects chosen from a set of n available
objects, when repetitions are not allowed.
Gold, Silver, Bronze
• Consider the top 3 winners in a race with 8
contestants. How many results are possible?
8!
8!
P 
  (8)(7)(6)  336
(8  3)! 5!
8
3
Or equivalently,
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
P  336
8
3
we simply enter:
8 nPr 3
Compare the 2 cases
Consider a club with 16 members:
• 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.
• 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:
P  3360
16
3
Case 2:
3360
C 
 560
6
16
3
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”.
Counting Combinations
• “Combinations of n objects, taken r at a time”
when repetitions are not allowed
n
r
P
n!
C 

r ! r !(n  r )!
n
r
Often read as “n, choose r"
n
Sometimes denoted as   , or Cn , r
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)?
n(hands with 4 spades, 1 non-spade)
n( possible 5-card hands)

C13,4C39,1
C52,5
(715)(39)

 0.01072929
2598960
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…
n
( x  y)   C x
n
k 0
n
k
nk
y
k
Remember “Pascal’s Triangle”?
Multinomial Coefficients
• “Ways to partition n objects into k groups”
when repetitions are not allowed
n


n!

n n

 1 2 nk   n1 !  n2 !
 nk !
Here the groups are of size n1, n2, …, and nk
such that n1 + n2 + … + nk = n.
Expanding a Multinomial
• Using the multinomial coefficients
 n  n1 n2 n3
( x  y  z)   
x y z
n1  n2  n3  n  n1 n2 n3 
n
In the expansion of the multinomial
( x  3 y  z)
8
Determine the coefficient of the x3y3z2 term.