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Chapter 5 Probability Theory. Rule 1 : The probability P(A) of any event A satisfies 0 < P(A) < 1 Rule 2 : If S is the sample space in a probability model,

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Presentation on theme: "Chapter 5 Probability Theory. Rule 1 : The probability P(A) of any event A satisfies 0 < P(A) < 1 Rule 2 : If S is the sample space in a probability model,"— Presentation transcript:

1 Chapter 5 Probability Theory

2 Rule 1 : The probability P(A) of any event A satisfies 0 < P(A) < 1 Rule 2 : If S is the sample space in a probability model, then P(S) = 1 Rule 3 : The complement of any event A is the event that A will not occur, written as A c. The complement rule states that P(A C ) = 1 - P(A) Rule 4 : (Addition rule for disjoint events) If A and B are disjoint, P(A or B) = P(A) + P(B) 5.1 General Probability Rules Rule 5 : (Multiplication rule) If A and B are independent events, P(A and B) = P(A) P(B) We have already met and used five rules in §4.2

3 General Addition Rule For any two events A and B, P(A or B) = P(A) + P(B)-P(A & B)

4 Example Two dice are thrown, what is the probability of 1)getting a total of 12 or an absolute difference of 4? B: A Total of 12, C An absolute difference of 4 P(B or C)=P(B)+P(C)=(1/36)+(4/36)=5/36 2)getting at least one “6” on both dice or an absolute difference of 4? D: At least one 6 P(C or D)=P(C)+P(D)-P(C&D) =(4/36)+(11/36)-(2/36)=13/36 3)getting an absolute difference of 4 or odd total? P(C or D)=P(C)+P(D)-P(C&D) =(4/36)+(18/36)-0=22/36

5 Caffeine Example Common sources of caffeine are coffee, tea and cola drinks. Suppose that: 55% of adults drink coffee, 25% of adults drink tea, 45% of adults drink cola 15% drink both coffee and tea, 5% drink all three beverages, 25% drink both coffee and cola And 5% drink only tea (a)Draw a Venn diagram marked with this notation. (b)What percent of adults drink only cola? (c)What percent drink none of these beverages?

6 The Binomial Setting: 1) There is a fixed number n of observations 2) The n observations are independent 3) Each observation falls into just one of two categories, which, for convenience, we call “success” and “failure”. 4) The probability of a success, call it p, is the same for each observation 5.2 The Binomial Distributions Example: Toss a fair coin 10 times, and count the number of heads which will appear. The random variable, X, is the number of heads that appear. The probability of a success is p = 0.5 This experiment satisfies the four requirements for the Binomial Setting, so this is a Binomial Distribution.

7 Binomial Distributions The distribution of the count X of successes in the binomial setting is called the binomial distribution with parameters n and p. The parameter n is the number of observations The parameter p is is the probability of a success on any one observation. The possible values of X are the whole numbers from 0 to n. We say that X is B(n,p)

8 Binomial Distributions Example: Imagine we have a big bunch of transistors, say roughly 10,000. We will say that 1000 of them are bad. Pull an SRS of size 10 without replacement. Let X be the amount of transistors which are bad in the SRS. Q: Is this the binomial setting? A: No. The first transistor has a 1000/10,000 = 0.1 chance of being bad. What about the second? The second has either a 1000/9,999 = 0.10001 or 999/9,999 chance = 0.09991 chance of being bad We can consider this a B(10,0.1)

9 Example In each of the following cases, decide whether or not a binomial setting is the appropriate model, and give your reasons. (a) Fifty students are taught about binomial distributions by a television program. After completing their study, all students take the same exam. The number of students who pass is counted. (b) A chemist repeats a solubility test 10 times on the same substance. Each test is conducted at temperature 10  higher than the previous test. She counts the number of times that the substance dissolves completely.

10 Example (cont.) In each of the following cases, decide whether or not a binomial setting is the appropriate model, and give your reasons. (c) A student studies binomial distributions using a computer-assisted instruction. After the initial instruction is completed, the computer presents 10 problems. The student solves each problem and enters the answer; the computer gives additional instruction between problems if the student’s answer is wrong. The number of problems that the student solves correctly is counted.

11 Binomial Probability Distribution Function n=sample size p=probability of ‘success’ k=number of ‘successes’ in sample (X = 0, 1, 2,..., n)

12 Example 1 Toss 1 Coin 4 times in a row. Note # Tails. What’s the Probability of 3 tails? This is a binomial setting with n=4, p=0.5 i.e. B(4,0.5) Using TI-83 Calculator under distributions menu Binompdf(4,0.5,3)=0.25

13 Example 1 Evans is concerned about a low retention rate for employees. On the basis of past experience, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employees chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year. Choosing 3 hourly employees a random, what is the probability that 1 of them will leave the company this year? Let: p =.10, n = 3, x = 1

14 Using the Binomial Probability Function = (3)(0.1)(0.81) = (3)(0.1)(0.81) =.243 =.243

15 Binomial Mean & Standard Deviation A: If a count X has the B(n,p) distribution, then :  X = np  X = np (1-p) Example: Refer back to Example 2 Expected Value E(x) = 3(.1) =.3 employees out of 3 E(x) = 3(.1) =.3 employees out of 3Variance Var(x) = 3(.1)(.9) =.27 Var(x) = 3(.1)(.9) =.27 Standard Deviation

16 The Normal approximation to Binomial distributions Suppose that a count X has the Binomial distribution with n trials and success probability p. When n is large, the distribution of X is approximately Normal, As a rule of thumb, we will use the Normal approximation when n and p satisfy

17 Examples Read Pages 329-330 Exercise 5.41, p. 334 The chance that Jodi will score 70% or lower is about 12.5% Slightly over 3% chance that Jodi will score 70% or lower.

18 5.3 The Poisson Distributions The Poisson Setting: 1.The number of successes that occur in any unit of measure is independent of the number of successes that occur in any non- overlapping unit of measure. 2.The probability that a success will occur in a unit of measure is the same for all units of equal size and is proportional to the size of the unit. 3.The probability that 2 or more successes will occur in a unit approaches 0 as the size of the unit becomes smaller.

19 Poisson Probability Function where: P(X=k) = probability of x occurrences in an interval µ = mean number of occurrences in an interval e = 2.71828

20 Using the Poisson Probability Function Patients arrive at the emergency room of Mercy Hospital at the average rate of 6 per hour on weekend evenings. What is the probability of 4 arrivals in 30 minutes on a weekend evening?  = 6/hour = 3/half-hour, x = 4

21 Example 5.15

22 Poisson Mean & Standard Deviation Mean = Variance = Standard Deviation =


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