Chapter 8 Counting Principles; Further Probability Topics

Chapter 8

Counting Principles;

Further Probability Topics

The Multiplication Principle; Permutations/Combinations

Counting Rules When we wish to know the number of all possible

outcomes for a sequence of events. Fundamental Counting Rule (Multiplication Principle) Permutation Rule Combination Rule


Fundamental Counting Rule (Multiplication Principle)

In a sequence of n events in which the first one has k possibilities and the second event has k and the third has k, and so forth, the total number of possibilities of the sequence will be

k1 · k2 · k3 · · · kn

where n is the number of events and k is the number of possible outcomes of each event


Example A quiz with four T/F questions. How many

possible answer keys?If n = 4 then

k1 = 2 k2 = 2 k3 = 2 k4 = 2

2·2·2·2 = 16

<TREE DIAGRAM>


A store manager wishes to display 8 different brands of shampoo in a row. How many ways can this be done?

If n = 8 then

k1 = 8 k2 = 7 k3 = 6 k4 = 5

k5 = 4 k6 = 3 k7 = 2 k8 = 1

8 ·7 ·6 ·5 ·4 ·3 ·2 ·1 = 40,320or 8! (factorial)

Factorials determine the number of ways in which objects or persons can be arranged in a line (Recall that 0! = 1).


Permutations The ordered arrangement of objects where “r” objects

are selected from a set of n distinct objects, i.e., 1st , 2nd , 3rd place out of 5 contestants.

nPr = n! .

(n - r)!


Combinations The arrangement of objects without regard to

order where “r” objects are selected from a set of n distinct objects (i.e., any 3 out of 5 contestants).


In a board of directors composed of 8 people, how many ways can 1 chief executive officer, 1 director, and 1 treasurer be selected? n = 8 r = 3 Need CEO, DIR., TRES.

Perm or Comb?

8P3 = 8! = 8! (8-3)! 5!

= 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 336 5 · 4 · 3 · 2 · 1

or40320 = 336 120


How many ways can a committee of 4 people be selected from a group of 10 people? n = 10 r = 4

Perm or Comb?

10C4 = 10! = 10! 4!(10-4)! 4!6!

= 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 4 · 3 · 2 · 1 · 6 · 5 · 4 · 3 · 2 · 1

or3628800 = 210 17280

Probability Distributions; Expected Value

Probability Distribution – any device (table, graph) used to specify all possible values of a variable along with its probabilities. Two types of probability distributions:

discrete random variables (r.v.) – only certain values e.g. whole numbers such as counts – people, cars, etc.

continuous random variables – continuum of values e.g. whole numbers and the numbers in between, such as measurements like height, weight, etc. random variable - a function that assigns a real number to

each outcome of an experiment

Probability Distributions; Expected Value The probabilities that a tutor sees 1, 2, 3, 4, or 5 students in any

one day are 0.10, 0.25, 0.25, 0.20, and 0.20 respectively

X 1 2 3 4 5

P(X) .10 .25 .25 .20 .20

1 2 3 4 5

P(x)

.3

.2

.1

Probability Distributions; Expected Value If a player rolls two dice and gets a sum of 2 or 12, she wins $20. If the person

gets a 7, she wins $5. The cost to play the game is $3. Find the expectation of the game.

Win LoseGain (X) $17 $2 -$3P(X) .0556 .1667 .7777

P(sum of 2) or P(sum of 12)= 1/36 + 1/36 = 2/36 = 1/18 = .0556

P(sum of 7)= 6/36 = 1/6 = .1667

P(o/w)= 1 - .0556 - .1667 = .7777

E(X) = 17(.0556) + 2(.1667) + -3(.7777) = -$1.05

Means that theoretically there will be an average loss of about a dollar

Probability Distributions; Expected Value A recent survey by an insurance company showed the

following probabilities for the number of automobiles each policyholder owned. Find the expected value.

# of autos, X 1 2 3 4 P(X) .4 .3 .2 .1

E(x) = X P(X) = 1(.4) + 2(.3) + 3(.2) + 4(.1) = 2

Binomial Probability

Binomial Experiment The same experiment is repeated several times

(a fixed number of times). There are only two possible outcomes:

Success Failure

The repeated trials are independent, so that the probability of success remains the same for each trial.


When a random variable can take on a large number of values with particular characteris-tics it is convenient to express the probability distribution in terms of a formula.


Binomial Probability Formula

P(X) can be written as b(x;n,p) is the same as nCx Mean (average) = = np Variance = 2 = np(1-p)

xnx ppxxn

nXP

)1(

!)!(

!)(

!)!(

!

xxn

n

Binomial Probability Binomial Distribution Notation

P(S) Probability of Success P(F) Probability of Failure p numerical probability of a success q= 1-p numerical probability of a failure P(S) = p P(F) = q=1-p n number of trials x number of successes 0 < x < n

Binomial Probability Using the Binomial Table

step 1: find page with sample size under consideration. step 2: find relevant value of p in column headings step 3: find desired value of x in second column from left. step 4: find probability at intersection of row x and column p.


Example: Given the following Binomial Experiment characteristics, use the Binomial Table to find the corresponding probabilities:

(a) n = 2 p= .3 X=1

b(1;2,.3) = .420

(b) n = 12 p = .90 X = 2

b(2;12,.9) = 0

(c) n = 20 p = .50 X = 10

b(10; 20, .5) = .176


EXAMPLE

If 20% of the people in a community use the emergency room at a hospital inone year, find these probabilities for a sample of 10 people:

(a) At most three used the emergency room(b) Exactly three used the emergency room(c) At least five used the emergency room


Given n = 10 and p = 0.20

(a) P(X<3) = P(x=0) + P(x=1) + P(x=2) + P(x=3) = 0.107 + 0.268 + 0.302 + 0.201 = 0.878

(b) P(X=3) = 0.201 (c) P(X > 5) = P(x=5) + P(x=6) + P(x=7) +P(x=8)+P(x=9)+P(x=10)

= .026 + .006 + .001 + .000 + .000 + .000 = .033

Rework (b) using binomial formula b(3:10,0.2)

= 10! ∙ 0.23 (1-.2)10-3

(10-3)!3! = 120(.2) 3 (.8) 7

= 0.201


Example

In a restaurant, a study found that 42% of allpatrons smoked. If the seating capacity of therestaurant is 80 people, how many seats should beavailable for smoking customers?


Given: P(smoker) = .42 n = 80

(average) = np = 80(.42) = 33.6

Therefore using mean as a “good” estimate, the restaurant should have about 34 seats available for smokers.

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Chapter 8 Counting Principles; Further Probability Topics