Independence and the Multiplication Rule Chapter 5 Section 3 Independence and the Multiplication Rule

Chapter 5 – Section 3 Learning objectives Understand independence Use the Multiplication Rule for independent events Compute at-least probabilities 1 2 3

Chapter 5 – Section 3 The Addition Rule shows how to compute “or” probabilities P(E or F)‏ under certain conditions The Multiplication Rule shows how to compute “and” probabilities P(E and F)‏ also under certain (different) conditions The Addition Rule shows how to compute “or” probabilities P(E or F)‏ under certain conditions

Chapter 5 – Section 3 Learning objectives Understand independence Use the Multiplication Rule for independent events Compute at-least probabilities 1 2 3

Chapter 5 – Section 3 The “disjoint” concept corresponds to “or” and the Addition Rule … disjoint events and adding probabilities The concept of independence corresponds to “and” and the Multiplication Rule … independent events and multiplying probabilities Basically, events E and F are independent if they do not affect each other The “disjoint” concept corresponds to “or” and the Addition Rule … disjoint events and adding probabilities The “disjoint” concept corresponds to “or” and the Addition Rule … disjoint events and adding probabilities The concept of independence corresponds to “and” and the Multiplication Rule … independent events and multiplying probabilities

Chapter 5 – Section 3 Definition of independence Events E and F are independent if the occurrence of E in a probability experiment does not affect the probability of event F Other ways of saying the same thing Knowing E does not give any additional information about F Knowing F does not give any additional information about E E and F are totally unrelated Definition of independence Events E and F are independent if the occurrence of E in a probability experiment does not affect the probability of event F

Chapter 5 – Section 3 Examples of independence Flipping a coin and getting a “tail” (event E) and choosing a card and getting the “seven of clubs” (event F)‏ Choosing one student at random from University A (event E) and choosing another student at random from University B (event F)‏ Choosing a card and having it be a heart (event E) and having it be a jack (event F)‏ Examples of independence Flipping a coin and getting a “tail” (event E) and choosing a card and getting the “seven of clubs” (event F)‏ Choosing one student at random from University A (event E) and choosing another student at random from University B (event F)‏ Examples of independence Flipping a coin and getting a “tail” (event E) and choosing a card and getting the “seven of clubs” (event F)‏ Examples of independence

Chapter 5 – Section 3 If the two events are not independent, then they are said to be dependent Dependent does not mean that they completely rely on each other … it just means that they are not independent of each other Dependent means that there is some kind of relationship between E and F – even if it is just a very small relationship If the two events are not independent, then they are said to be dependent If the two events are not independent, then they are said to be dependent Dependent does not mean that they completely rely on each other … it just means that they are not independent of each other

Chapter 5 – Section 3 Examples of dependence Examples of dependence Whether Jack has brought an umbrella (event E) and whether his roommate Joe has brought an umbrella (event F)‏ Choosing a card and having it be a red card (event E) and having it be a heart (event F)‏ The number of people at a party (event E) and the noise level at the party (event F)‏ Examples of dependence Whether Jack has brought an umbrella (event E) and whether his roommate Joe has brought an umbrella (event F)‏ Choosing a card and having it be a red card (event E) and having it be a heart (event F)‏ Examples of dependence Whether Jack has brought an umbrella (event E) and whether his roommate Joe has brought an umbrella (event F)‏ Examples of dependence

Chapter 5 – Section 3 What’s the difference between disjoint events and independent events? Disjoint events can never be independent Consider two events E and F that are disjoint Let’s say that event E has occurred Then we know that event F cannot have occurred Knowing information about event E has told us much information about event F Thus E and F are not independent What’s the difference between disjoint events and independent events? Disjoint events can never be independent Consider two events E and F that are disjoint Let’s say that event E has occurred Then we know that event F cannot have occurred Knowing information about event E has told us much information about event F What’s the difference between disjoint events and independent events? Disjoint events can never be independent Consider two events E and F that are disjoint Let’s say that event E has occurred What’s the difference between disjoint events and independent events?

Chapter 5 – Section 3 Learning objectives Understand independence Use the Multiplication Rule for independent events Compute at-least probabilities 1 2 3

Chapter 5 – Section 3 The Multiplication Rule for independent events states that P(E and F) = P(E) • P(F)‏ Thus we can find P(E and F) if we know P(E) and P(F)‏

Chapter 5 – Section 3 This is also true for more than two independent events If E, F, G, … are all independent (none of them have any effects on any other), then P(E and F and G and …)‏ = P(E) • P(F) • P(G) • … This is also true for more than two independent events

Chapter 5 – Section 3 Example P(E and F)‏ P(E and F) = 1/8 Example E is the event “draw a card and get a diamond” F is the event “toss a coin and get a head” E and F are independent P(E and F)‏ We first draw a card … with probability 1/4 we get a diamond When we toss a coin, half of the time we will then get a head, or half of the 1/4 probability, or 1/8 altogether P(E and F) = 1/8 Example E is the event “draw a card and get a diamond” F is the event “toss a coin and get a head” E and F are independent

P(E and F) = P(E) • P(F) = 1/4 • 1/4 = 1/16 Chapter 5 – Section 3 Another example E is the event “draw a card and get a diamond” Replace the card into the deck F is the event “draw a second card and get a spade” E and F are independent P(E and F)‏ P(E and F) = P(E) • P(F) = 1/4 • 1/4 = 1/16 Another example E is the event “draw a card and get a diamond” Replace the card into the deck F is the event “draw a second card and get a spade” E and F are independent

Chapter 5 – Section 3 The previous example slightly modified E is the event “draw a card and get a diamond” Do not replace the card into the deck F is the event “draw a second card and get a spade” E and F are not independent Why aren’t E and F independent? After we draw a diamond, then 13 out of the remaining 51 cards are spades … so knowing that we took a diamond out of the deck changes the probability for drawing a spade The previous example slightly modified E is the event “draw a card and get a diamond” Do not replace the card into the deck F is the event “draw a second card and get a spade” E and F are not independent

Chapter 5 – Section 3 Learning objectives Understand independence Use the Multiplication Rule for independent events Compute at-least probabilities 1 2 3

Chapter 5 – Section 3 There are probability problems which are stated: What is the probability that "at least" … For example At least 1 means 1 or 2 or 3 or 4 or … At least 5 means 5 or 6 or 7 or 8 or … These calculations can be very long and tedious The probability of at least 1 = the probability of 1 + the probability of 2 + the probability of 3 + the probability of 4 + … There are probability problems which are stated: What is the probability that "at least" … There are probability problems which are stated: What is the probability that "at least" … For example At least 1 means 1 or 2 or 3 or 4 or … At least 5 means 5 or 6 or 7 or 8 or …

Chapter 5 – Section 3 There is a much quicker way using the Complement Rule Assume that we are counting something E = “at least one” and we wish to compute P(E)‏ Ec = the complement of E, when E does not happen Ec = “exactly zero” Often it is easier to compute P(Ec) first, and then compute P(E) as 1 – P(Ec)‏ There is a much quicker way using the Complement Rule Assume that we are counting something E = “at least one” and we wish to compute P(E)‏ Ec = the complement of E, when E does not happen Ec = “exactly zero” There is a much quicker way using the Complement Rule Assume that we are counting something E = “at least one” and we wish to compute P(E)‏ There is a much quicker way using the Complement Rule There is a much quicker way using the Complement Rule Assume that we are counting something E = “at least one” and we wish to compute P(E)‏ Ec = the complement of E, when E does not happen

Chapter 5 – Section 3 Example We flip a coin 5 times … what is the probability that we get at least 1 head? E = {at least one head} Ec = {no heads} = {all tails} Ec consists of 5 events … tails on the first flip, tails on the second flip, … tails on the fifth flip These 5 events are independent P(Ec) = 1/2 • 1/2 • 1/2 • 1/2 • 1/2 = 1/32 Thus P(E) = 1 – 1/32 = 31/32 Example We flip a coin 5 times … what is the probability that we get at least 1 head? E = {at least one head} Ec = {no heads} = {all tails}

Summary: Chapter 5 – Section 3 The Multiplication Rule applies to independent events, the probabilities are multiplied to calculate an “and” probability Probabilities obey many different rules Probabilities must be between 0 and 1 The sum of the probabilities for all the outcomes must be 1 The Complement Rule The Addition Rule (and the General Addition Rule)‏ The Multiplication Rule

Examples A manufacturer of exercise equipment knows that 10% of their products are defective. They also know that only 30% of their customers will actually use the equipment in the first year after it is purchased. If there is a one-year warranty on the equipment, what proportion of the customers will actually make a valid warranty claim?

(3%)

Examples The No Child Left Behind (NCLB) Act of 2001 mandates that by the year 2014, each public school must achieve 100% proficiency. (Source: www.ed.gov/nclb) That is, every student must pass a proficiency exam or the school faces serious sanctions. Suppose that students in a particular school are all very competent. For each individual, the probability that they will pass the proficiency exam (independent of their classmates) is 0.99. a. If there are 500 students in the school, what is the probability that everyone will pass the exam? b. What is the probability that at least one of the 500 students will not pass the proficiency exam (and the school will not meet the mandate)?

(0.00657) (0.99343)

Examples Repeat problem 1 for a school of 1000 very competent students (with individual success probabilities of 0.99).

(0.00004; 0.99996)

Examples In a more typical school, suppose that for each individual student, the probability that they will pass the proficiency exam is 0.85. a. If there are 500 students in the school, what is the probability that everyone will pass the exam? Is this event likely? b. What is the probability that at least one of the 500 students will not pass the proficiency exam (and the school will not meet the mandate)? Is this event likely?

(5.122×10-36; no) (1-5.122×10-36; yes)

Examples The following table gives the number (in millions) of men and women over the age of 24 at each level of hightest educational attainment. (Source: U.S. Department of Commerce, Census Bureau, Current Population Survey, March 2004.) a. What is the probability that two randomly selected females over the age of 24 are both college graduates? b. What is the probability that two randomly selected people over the age of 24 are both women? c. Suppose four people over the age of 24 are randomly selected, what is the probability that at least one will be a college graduate? d. Suppose three people over the age of 24 are randomly selected, what is the probability that at least one will not have attended college?

(0.1250) (0.2710) (0.8334) (0.8500)