Chapter 4 Section 4.1 Statistics Mr. zboril | Milford PEP

Chapter 4 Section 4.1 Probability You are important to me – don’t ever think otherwise! This Photo y Unknown Author is licensed under CC BY-SA

Section 4.1 Probability Focus Points Assign probabilities to events. Explain how the law of large numbers relates to relative frequencies. Apply basic rules of probability in everyday life. Relationship between statistics and probability.

Section 4.1 Probability When we use probability in a statement, we’re using a number between 0 and 1 to indicate the likelihood of an event. Probability is a numerical measure between 0 and 1 that describes the likelihood that an event will occur. Closer to 1 – more likely to occur Closer to 0 – less likely to occur

Section 4.1 Probability of Event P(A) P(A) reads “P of A” is the probability of event A P(A) = 1, event A is certain to occur P(A) = 0, event A is certain not to occur Closer to 1 – more likely to occur Closer to 0 – less likely to occur

Section 4.1 Example 1 Probability Assignment A probability assignment based on intuition incorporates past experience, judgement, or opinion to estimate the likelihood of an event. A sports announcer states, “Shelia has a 90% chance of breaking the world record in the 100-yard dash.” This probability was assigned based on intuition.

Section 4.1 Example 1 Probability Assignment Ricky figures if he guesses on a true-false question, the probability of getting it right is 0.50. This probability assignment is based on equally likely outcomes. There are 2 likely outcomes and only 1 is correct. 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑦 𝑜𝑓 𝑒𝑣𝑒𝑛𝑡 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑎𝑛𝑠𝑤𝑒𝑟 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 1 2 =0.50

Section 4.1 Example 1 Probability Assignment Ricky figures if he guesses on a true-false question, the probability of getting it right is 0.50. In real life, unfortunately, whenever you are faced with a 50-50 chance of choosing the right answer, you will pick the wrong choice 90% of the time. A close cousin to Murphy’s Law, this is known as the ‘50-50-90 Rule’. 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑦 𝑜𝑓 𝑒𝑣𝑒𝑛𝑡 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑎𝑛𝑠𝑤𝑒𝑟 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 1 2 =0.50

Section 4.1 Example 1 Probability Assignment The National Homeowners Insurance Association claims that the probability of a hurricane making landfall in the United States is .3 based on studying the tracks of hurricanes for the last 100 years and finding 30 years where a storm hit the continental United States. This probability assignment is based on relative frequency. 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑦 𝑜𝑓 𝑒𝑣𝑒𝑛𝑡=𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦= 𝑓 𝑛 = 30 100 =0.30

Section 4.1 Probability Assignment There are three ways to assign probability: intuition, relative frequency, and – when outcomes are equally likely - a formula. To know which method to use depends on the information you have. Probabilities are numbers between 0 and 1 – don’t assign probabilities outside this range.

Section 4.1 Law of Large Numbers Relative frequency is commonly used for probability assignment. The underlying assumption, of course, is if events occurred a certain percentage of times in the past, they will occur about the same percentage of times in the future. Law of Large Numbers In the long run, as the sample size becomes larger and larger, the relative frequency of outcomes get closer and closer to the theoretical (or actual) value.

Section 4.1 Law of Large Numbers Think about it, why would insurance companies write policies on homes being built in hurricane or earthquake-prone areas? Not only would they write the policy but you will have insurers competing for the policy! They know that the likelihood of the event actually happening is far lower than you would otherwise think. How do they know this? Because they employ mathematicians and statisticians to do the research, assign the risk, and calculate the highest policy they can charge while remaining both profitable and competitive. This also holds true for health and auto insurance along with casinos.

Section 4.1 Statistical Experiments To determine the possible outcomes, we need to define a statistical experiment. A statistical experiment or statistical observation can be thought of as any random activity that results in a definite outcome. A sample event is one particular outcome of a statistical experiment. The set of all sample events constitutes the sample space of an experiment.

Section 4.1 Example 2 on Page 135 Human eye color is determined by a single pair of genes – one from both mom and dad. Brown eye color (or B) is dominant over blue eye color (𝓁). Look at Table 4-1. What is the probability the child will have blue eyes? Father Mother B 𝓁 B BB B𝓁 𝓁 𝓁B 𝓁𝓁

Section 4.1 Example 2 on Page 135 Since all four outcomes are equally likely, we will use the equally likely outcomes formula. Father Mother B 𝓁 B BB B𝓁 𝓁 𝓁B 𝓁𝓁 𝑃(𝑏𝑙𝑢𝑒 𝑒𝑦𝑒𝑠)= 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 1 4 =0.25 𝑃(𝑏𝑟𝑜𝑤𝑛 𝑒𝑦𝑒𝑠)= 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 3 4 =0.75

Section 4.1 Compliment of an Event The complement of event A is the event that A does not occur. Ac designates the complement of event A. P(A) + P(Ac) = 1 P(event A does not occur) = P(Ac) = 1 – P(A) Example 3 pg. 136 The probability that a college student who has not received the flu shot will get the flu is 0.45. What is the probability that a student will not get the flu if he or she does not get the flu shot?

Section 4.1 Compliment of an Event The complement of event A is the event that A does not occur. Ac designates the complement of event A. P(A) + P(Ac) = 1 P(event A does not occur) = P(Ac) = 1 – P(A) Example 3 pg. 136 P(will get the flu) = 0.45 P(will not get the flue = 1 - .45 = .55

Section 4.1 Questions?

Chapter 4 Section 4.2 Statistics Mr. zboril | Milford PEP This Photo by Unknown Author is licensed under CC BY-SA

Chapter 4 Section 4.2 Some Probability Rules You are important to me – don’t ever think otherwise!

Chapter 4 Section 4.2 Some Probability Rules Focus Points Compute probabilities of compound events. Compute probabilities involving independent events or mutually exclusive events Use survey results to compute conditional probabilities

Chapter 4 Section 4.2 Some Probability Rules You roll two dice. What is the probability you roll a 5 on each die? There are 6 balls – identical in every way except color. There are 3 green balls, 2 blue balls, and 1 red ball. You draw two balls at random. Without replacing the first ball, what is the probability both balls will be green?

Chapter 4 Section 4.2 Some Probability Rules It seems these two problems are the same, but they are not. Two events do occur together. For the dice, P(5 on the 1st die and 5 on the 2nd) For the balls, P(green on the 1st draw and green on the 2nd draw)

Chapter 4 Section 4.2 Some Probability Rules Since the outcome of the 1st die does not affect the probability of a 5 on the 2nd die, the events are independent. Two events are independent if the occurrence or nonoccurrence of one event does not change the probability that the other event will occur. Multiplication rule for independent events P(A and B) = P(A) · P(B) 1 6 x 1 6 = 1 / 36

Chapter 4 Section 4.2 Some Probability Rules The probability of drawing a green ball on the 1st draw is 3 / 6 since there six balls and three of them are green. HOWEVER… ….the probability of drawing a green on the 2nd draw is 2 / 5 since only 5 balls remain and 2 of them are green. Since the outcome of the 1st draw has a direct effect on the 2nd draw, the events are dependent.

Chapter 4 Section 4.2 Some Probability Rules Since these events are dependent, we need to use a different formula and terminology. The notation P(A, given B) denotes the probability that event A will occur if event B occurs. The standard notation for P(A, given B) is P(A | B).

Chapter 4 Section 4.2 Some Probability Rules If events A and B are dependent events, then P(A) ≠ P(A, given B) because the occurrence of event B has changed the probability that event A will occur. General multiplication rule for any events P(A and B) = P(A) · P(B | A) P(A and B) = P(B) · P(A | B)

Chapter 4 Example 5 Page 145 Example 5 Determine the probability of drawing two green balls. Multiplication Rule : The probability of a green ball on the 1st draw is 3 / 6, but 2 / 5 on the 2nd draw. 3 6 · 2 5 = 6 30 = 1 5 Sample Space Method : see bottom of page 145

Chapter 4 Example 5 Page 145 Multiplication Rule : The probability of a green ball on the 1st draw is 3 / 6, but 2 / 5 on the 2nd draw. P(A and B) = P(A) · P(B | A) P(green on 1st draw and green on 2nd draw.) = P(green on 1st) x P(green on 2nd | green on 1st) 3 6 · 2 5 = 6 30 = 1 5

Chapter 4 How to Use Multiplication Rules First determine whether A and B are independent. If P(A) = P(A | B), then events are independent. If A and B are independent events, P(A and B) = P(A) · P(B). If A and B are any events, P(A and B) = P(A) · P(B | A) P(A and B) = P(B) · P(A | B) Multiplication Rule : The probability of a green ball on the 1st draw is 3 / 6, but 2 / 5 on the 2nd draw. 3 6 · 2 5 = 6 30 = 1 5 Sample Space Method : see bottom of page 145

Chapter 4 How to Use Multiplication Rules Guided Exercise 4 p. 146 Guided Exercise 5 p. 147

Chapter 4 Section 4.2 Addition Rules The difference between the probabilities of events A and B occurring compared to events A or B occurring can be shown on the bottom of page 147. Example 6 Statistics Class with 31 students. See Figure 4-5 at bottom of pg. 148. Select one student at random. What is the probability the student is a freshman or a sophomore? 15 of the 31 students are freshman

Chapter 4 Section 4.2 Addition Rules – Mutually Exclusive Select one student at random. What is the probability the student is a freshman or a sophomore? 15 of the 31 students are freshman 8 of the 31 students are sophomores 15 31 + 8 31 = 23 31 𝑜𝑟 0.742 This event is mutually exclusive since if a student is a freshman, they can’t be a sophomore. Addition rule for mutually exclusive events A and B. P(A or B) = P(A) + P(B)

Chapter 4 Section 4.2 Addition Rules – General Events Select one student at random. What is the probability the student is a male or a sophomore? 14 of the 31 students are male 8 of the 31 students are sophomores We can’t simply add these up like the previous problem since some students are both male and sophomores. 5 of the 31 students are male and sophomores Addition rule for general events A and B. P(A or B) = P(A) + P(B) – P(A and B)

Chapter 4 Section 4.2 Addition Rules Examples Guided Exercise 7 Pg.150 Guided Exercise 8 Pg.151 Addition rule for mutually exclusive events A and B. P(A or B) = P(A) + P(B) Addition rule for general events A and B. P(A or B) = P(A) + P(B) – P(A and B)

Chapter 4 Section 4.2 Surveys Simple tally survey is a popular method of sampling. A contingency table is a set of are rows and columns where survey question responses can be recorded. (See Table 4-2 on pg. 152).

Chapter 4 Section 4.2 Formula Review Multiplication rule for independent events P(A and B) = P(A) · P(B) 1 6 x 1 6 = 1 / 36 General multiplication rule for any events P(A and B) = P(A) · P(B | A) P(A and B) = P(B) · P(A | B) This Photo y Unknown Author is licensed under CC BY-SA

Chapter 4 Section 4.2 Formula Review When you multiply fractions and decimals <1 together the product is always a smaller number. That is why when we combine probabilities with an AND, such as the probability of P(A and B), we MULTIPLY the fractions or decimals together. The P(A and B) is always going to be less likely than the probability of P(A) or P(B) alone. Multiplication rule for independent events P(A and B) = P(A) · P(B) 1 6 x 1 6 = 1 / 36 General multiplication rule for any events P(A and B) = P(A) · P(B | A) P(A and B) = P(B) · P(A | B) This Photo y Unknown Author is licensed under CC BY-SA

Chapter 4 Section 4.2 Formula Review Problem #19 p. 157 We are drawing two cards without replacing the first card. Since the 2nd draw is dependent on the first draw P(B | A) we use the General Multiplication rule. General multiplication rule for any events P(A and B) = P(A) · P(B | A) P(A and B) = P(B) · P(A | B) b) The likelihood the 1st card will be an Ace is 4 52 . P(A) = 4 52 . The card is NOT replaced so the prob. the 2nd card will be a King is 4 51 . P(B | A) = 4 51 . 4 52 𝑥 4 51 = 16 2652 OR .006. This Photo y Unknown Author is licensed under CC BY-SA

Chapter 4 Section 4.2 Formula Review It is an unlikely event that you will draw an Ace on the 1st draw – 4 out of 52. Should it be any surprise, then, the probability of drawing two specific cards in two draws diminishes to 4/663? When you multiply fractions together the answer will always be less – sometimes far less. “AND” events involve MULTIPLYING fractions. “OR” events involve ADDITION This Photo y Unknown Author is licensed under CC BY-SA

Chapter 4 Section 4.2 Bayes’s Theorem Bayes Theorem is used to determine the probability that Event A occurs given Event B WHEN WE KNOW the probability of Event B | A and B | Ac. A Ac B and Ac B B and A

Chapter 4 Section 4.2 Bayes Theorem Bayes Theorem is used to determine the probability that Event A occurs given Event B when we know the probability of Event B | A and B | Ac. Bayes’s Theorem P(A | B) = 𝑃 𝐵 𝐴 𝑃(𝐴) 𝑃 𝐵 𝐴 𝑃 𝐴 +𝑃 𝐵 𝐴𝑐 𝑃 𝐴𝑐

Section 4.2 Questions?