Chapter 3 Lecture

Probability To make decisions big and small it is often necessary to make an educated guess at the outcome. For instance, someone may choose their driving route to work based on their guess of traffic conditions. Or college students may choose a major based on their best guess at the career path that will be most fulfilling. Probability studies the likelihood of an event to occur considering all possible outcomes. In this chapter we will:  Learn how to calculate basic probabilities  Study the complement of an event  Learn the addition and multiplication rules of probability  Study contingency tables and probability trees  Learn how to count the number of possible outcomes for multiple events and when using permutations and combinations

Sample Space and Event: sample space - the set of all possible outcomes, denoted S event - a subset of the sample space, denoted by a capital letter such as A Example : In a standard deck of 52 cards, consider drawing one card at random in the hopes of drawing a face card. A list of all 52 cards is the sample space, and a list of only those which are face cards is the event. Here is the sample space, S . And here is the event, A .

Problem #2 : Calculate the probability of the event. Write as both a fraction and as a decimal rounded to 3 places. a) Drawing a red ace from a deck of 52 cards. b) A single die is rolled and the number is less than 9. c) In a family of three children exactly two are girls. To find the probability that several independent events each occur in order, take the product of their individual probabilities. d) When a die is rolled two times, the first number is 3 and the second is even. The Complement of an Event: complement of event A - all outcomes for which A does not occur, denoted by A It follows that P ( A ) = 1 − P ( A ) Example : In a family of three children, say event A is to have exactly two girls. a) Which outcomes are in A ? BBB BBG BGB GBB GGG b) What is P ( A ) ? P ( A )= 5 8 = 0.625

Example : In a deck of 52 cards, four cards are drawn. After each draw the card is placed back into the deck and they are shuffled before drawing again. What is the probability that at least one ace is drawn? Round to 3 decimal places. Here A = at least one aceis drawn This includes cases where exactly one ace is drawn, exactly two aces are drawn, exactly three are drawn, and exactly four are drawn. So, A = noaces aredrawn To calculate P ( A ) , find the probability of not drawing an ace on any of the four draws. P ( A )= ( no ace 1 st ) ∙ ( noace 2 nd ) ∙ ( noace 3 rd ) ∙ ( noace 4 th ) P ( A )= 48 52 ∙ 48 52 ∙ 48 52 ∙ 48 52 = 0.726 Then find P ( A ) = 1 − P ( A ) P ( A ) = 1 − 0.726 P ( A ) = 0.274 Problem #3 : A coin is tossed 5 times. Find the probability of getting at least one tail.

A helpful tool to visualize P ( A ∧ B ) is a Venn diagram. In the last example, we were told P ( A ∧ B ) = 0.06 or 6%. On the diagram, this would be the area taken up by the overlap of regions A and B. The problem stated that the outcome was already known to lie inside region A . So, to find the probability that B would also occur take the area of the overlap and divide it by the total area of A . OR Events and the Addition Rule: What if instead it is desired to know the probability of either event to occur? The Venn diagram will be helpful again. This would mean the outcome lies anywhere in A or B. P ( A ∨ B ) = P ( A ) + P ( B ) − P ( A ∧ B )

Example : A single card is drawn from a deck of 52 cards. What is the probability that it is either an ace or is black? Round to 3 decimal places. A = ace B = ¿ P ( A ∨ B ) = P ( A ) + P ( B ) − P ( A ∧ B ) P ( A ∨ B ) = 4 52 + 26 52 − 2 52 P ( A ∨ B ) = 0.538 Problem #5 : A box contains 3 glazed doughnuts, 4 jelly doughnuts, and 5 chocolate doughnuts. If a doughnut is selected at random, what is the probability that it is either glazed or chocolate? mutually exclusive (disjoint) events - the probability that both occur is zero In probability notation, P ( A ∧ B ) = 0 . This Venn diagram where there is no overlap depicts mutually exclusive events A and B .

P ( F | VH ) = Number very happy females Totalvery happy = 149 370 = 0.403 The GIVEN formula can also be used, although not necessary. P ( F | VH ) = P ( F ∧ VH ) P ( VH ) = 149 603 370 603 = 149 603 ∙ 603 370 = 149 370 = 0.403 Problem #6: From the same table above, answer these questions. a) What is the probability someone from the survey is not happy or female? b) What is the probability that someone is very happy given that they are female? Problem #7 : Say 13% of students are computer science majors (event A), 15% have taken discrete mathematics (event B), and 84% of students are not computer science majors and have not taken discrete mathematics. What is the probability that a student is a not a computer science major but has taken discrete mathematics? Before trying to answer, complete this contingency table. A A Total B 0.15 B 0.84 Total 0.13

Probability Trees: For complicated conditional probabilities, when event A influences the probability of event B , a tree diagram is helpful to visualize the relationships. Example : A college professor knows that of the students who attend the final review session, 90% pass the final exam. Of those who do not attend the review, only 65% pass the final. This semester, the professor estimates that 80% of the students will attend the final review session. For each student, let event R be “attend the review” and event P be “pass the final”. a) Create a probability tree. b) What percent of students passed the final exam? P ( Pass ) = P ( R ∧ P ) + P ( R ∧ P ) = 0.72 + 0.13 = 0.85

How many different kinds of paint can be made if you can select one color, one type, one texture, and one use? Example : The manager of a department store chain wishes to make four-digit id cards for her employees. a) How many different cards can be made if she uses the digits 1, 2, 3, 4, 5, 6 where repeating numbers is permitted. ( 1 st digit ) ∙ ( 2 nd digit ) ∙ ( 3 rd digit ) ∙ ( 4 th digit ) = 6 ∙ 6 ∙ 6 ∙ 6 = 1296 b) What if repeating any number is not permitted? ( 1 st digit ) ∙ ( 2 nd digit ) ∙ ( 3 rd digit ) ∙ ( 4 th digit ) = 6 ∙ 5 ∙ 4 ∙ 3 = 360 permutation - an arrangement in which different sequences of the same items are counted separately For example, with the letters {a, b, c} , the arrangements abc , bac , cab , acb , bca , and cba are all counted separately combination - an arrangement in which different sequences of the same items are not counted separately For example, with letters {a, b, c} , the arrangements abc , bac , cab , acb , bca , and cba are all considered to be the same combination. Permutations and combinations often result in a large number of possible outcomes. Here are the formulas for r objects taken from a total of n choices. P ( n,r ) = n! ( n − r ) ! C ( n,r ) = n! r ! ( n − r ) ! Example :

A television news director wishes to use 3 news stories on an evening show. One story will be the lead story, one will be the second story, and the last will be a closing story. If the director has a total of 8 stories to choose from, how many possible ways can the news program be set up? This is a permutation since the order matters. Use n = 8 and r = 3 . P ( 8 , 3 ) = 8 ! ( 8 − 3 ) ! = 8 ! 5 ! = 8 ∙ 7 ∙ 6 = 336 Example : A online journal has received 12 books to review. The editor decides that they can use 2 reviews in their next publication. How many different ways can these 3 be selected? This is a combination since order does not matter. C ( 12,2 ) = 12 ! 2 ! ( 12 − 2 ) ! = 12 ! 2 !∙ 10 ! = 12 ∙ 11 2 = 66 Problem #10 : In a club there are 7 women and 5 men. A committee of 3 women and 2 men is to be chosen. How many different ways are there to choose the committee members? Problem #11 : A school musical director can select 2 musical plays to present next year. One will be presented in the fall, and one will be presented in the spring. If she has 9 to pick from, how many different possibilities are there?

Wrap Up After studying the many varieties and applications of probability, which of these topics seems the most clear to you? You can choose more than one or none of these.  complement of event A  AND events  OR events  contingency tables  probability trees  permutations and combinations Which topic is the most confusing to you? Could you articulate why?  complement of event A  AND events  OR events  contingency tables  probability trees  permutations and combinations