1. Chapter 8 Counting Principles; Further Probability Topics

2. 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

3. The Multiplication Principle; Permutations/Combinations 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 k 1 · k 2 · k 3 · · · k n where n is the number of events and k is the number of possible outcomes of each event

4. The Multiplication Principle; Permutations/Combinations Example A quiz with four T/F questions. How many possible answer keys? If n = 4 then k 1 = 2 k 2 = 2 k 3 = 2k 4 = 2  2·2·2·2 = 16 

5. The Multiplication Principle; Permutations/Combinations 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 k 1 = 8 k 2 = 7 k 3 = 6k 4 = 5 k 5 = 4 k 6 = 3 k 7 = 2k 8 = 1  8 ·7 ·6 ·5 ·4 ·3 ·2 ·1 = 40,320 or 8! (factorial) Factorials determine the number of ways in which objects or persons can be arranged in a line (Recall that 0! = 1).

6. The Multiplication Principle; Permutations/Combinations Permutations The ordered arrangement of objects where “r” objects are selected from a set of n distinct objects, i.e., 1 st, 2 nd, 3 rd place out of 5 contestants.  n P r = n!. (n - r)!

7. The Multiplication Principle; Permutations/Combinations 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).

8. The Multiplication Principle; Permutations/Combinations 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 = 8r = 3 Need CEO, DIR., TRES. Perm or Comb?  8 P 3 = 8! = 8! (8-3)! 5! = 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 336 5 · 4 · 3 · 2 · 1 or 40320 = 336 120

9. The Multiplication Principle; Permutations/Combinations How many ways can a committee of 4 people be selected from a group of 10 people? n = 10r = 4 Perm or Comb?  10 C 4 = 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 or 3628800 = 210 17280

10. 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

11. 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

12. 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. WinLose Gain (X) $17 $2 -$3 P(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

13. 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, X1234 P(X).4.3.2.1 E(x) =  X  P(X) = 1(.4) + 2(.3) + 3(.2) + 4(.1) = 2

14. 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.

15. Binomial Probability 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.

16. Binomial Probability Binomial Probability Formula P(X) can be written as b(x;n,p) is the same as n C x Mean (average) =  = np Variance =  2 = np(1-p)

17. Binomial Probability Binomial Distribution Notation P(S) Probability of Success P(F)Probability of Failure pnumerical probability of a success q= 1-p numerical probability of a failure P(S) = p P(F) = q=1-p nnumber of trials xnumber of successes 0 < x < n

18. 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.

19. Binomial Probability 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

20. Binomial Probability EXAMPLE If 20% of the people in a community use the emergency room at a hospital in one 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

21. Binomial Probability 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! ∙ 0.2 3 (1-.2) 10-3 (10-3)!3! = 120(.2) 3 (.8) 7 = 0.201

22. Binomial Probability Example In a restaurant, a study found that 42% of all patrons smoked. If the seating capacity of the restaurant is 80 people, how many seats should be available for smoking customers?

23. Binomial Probability 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.