Term Definition Example Random Phenomena

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1 UNIT VI STUDY GUIDE Probabilities Course Learning Outcomes for Unit VI Upon completion of this unit, students should be able to: 1. Apply mathematical principles used in real-world situations. 1.1 Demonstrate how probability theory is used in genetics. 1.2 Demonstrate the relationship between probability and odds. 1.3 Evaluate the relationship between the probability of an event and the probability of its complement. 6. Apply methods for computing probabilities. 6.1 Calculate probabilities by counting outcomes in a sample space. 6.2 Apply counting formulas to compute probabilities. 6.3 Calculate the probability of the union of two events. 6.4 Apply complement and union formulas to compute the probability of an event. 6.5 Demonstrate how to compute conditional probability. 6.6 Calculate the probability of the intersection of two events. 6.7 Construct probability trees to compute conditional probabilities. 6.8 Identify the difference between dependent and independent events. Reading Assignment Chapter 13: Probability: What Are the Chances? Section 13.1: The Basics of Probability Theory, pp Section 13.2: Complements and Unions of Events, pp Section 13.3: Conditional Probability and Intersections of Events, pp Unit Lesson Probability is the likeliness that an event will occur. The local weatherman may use the concepts of probability to determine if it is going to rain. Although we can never be certain of the weather, probability can give us an idea of how often an event could happen Sample Spaces and Events: In order to calculate the probability, you must first understand the concepts related to probability. The table below defines the key terms that are important when solving probability problems. The example in the table refers to the action of flipping a coin. Term Definition Example Random Phenomena Experiment Outcomes An occurrence in which the result is never known with certainty. An observation of random phenomena. The different possible results of the experiment. Flipping a coin Results of flipping a coin Heads and Tails MAT 1301, Liberal Arts Math 1

2 Sample Space Event The set of all possible outcomes of the experiment. The sample space is sometimes denoted with brackets, { }. A subset of the sample space. We usually find the probability of an event. {heads, tails} Heads or Tails (We can be asked to find the probability of heads appearing or the probability of tails appearing when we flip a coin.) Write the event as a set of outcomes. Event: We flip three coins and obtain more (H)eads than (T)ails. Recall, that each coin can be flipped so that either a head or tail appears. To begin, list all possible outcomes of flipping three coins. Then, list the set in brackets { }, and separate each outcome by a comma. For example, if the first coin is flipped and is a head, the second coin is flipped and is a head, and the third coin is flipped and is a head, the outcome would be HHH. We can go through this process until all outcomes are listed: Next, the outcomes will be selected where the number of heads is more than the number of tails. {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Therefore, the set of outcomes for flipping three coins and obtaining more heads than tails is as follows: MAT 1301, Liberal Arts Math 2

3 Empirical information found from an experiment can be used to calculate probabilities. UNIT x STUDY This means GUIDE that the results from an experiment will be used to assign probabilities to specific events. The probability of an event, P(E), can easily be calculated by using the empirical assignment of probabilities formula. Empirical Assignment of Probabilities If E is an event and an experiment is performed several times, then the probability of E is estimated as follows: P(E) = This is sometimes called the relative frequency of E. the number of times E occurs the number of times the experiment is performed The residents of a small town and the surrounding area are divided over the proposed construction of a sprint car racetrack in the town. Information on the opinion of the residents is represented in the table. Support Racetrack Live in Town 1,512 2,268 Live in Surrounding Area 3,528 1,764 Oppose Racetrack If a newspaper reporter randomly selects a person to interview from these people, what is the probability that the person supports the racetrack? The empirical assignment of probabilities is used to solve this problem. The event is a person who supports the racetrack. Therefore, we will label the formula as follows: Counting and Probability P(supporting racetrack) = P(supporting racetrack) = the number of people who support the racetrack the number of times the experiment is performed 1, ,528 1, , , ,764 P(supporting racetrack) = 5,040 = 0.56 = 56% 9,072 Theoretical information can be used to calculate the probability by identifying the number of times an event occurs and the number of all possible outcomes in the experiment. The following examples require us to find the sample space of event. If all possible outcomes within the sample space have the same probability, then the formula for calculating probability when outcomes are equally likely will be used. Calculating Probability When Outcomes Are Equally Likely If E is an event in a sample space S with all equally likely outcomes, then the probability of E is given by the formula: P(E) = number of times the event occurs number of all possible outcomes = n(e) n(s) MAT 1301, Liberal Arts Math 3

4 Use the spinner to find the probability of the event. Event: Blue appears twice in two spins of the spinner. First, all possible outcomes of spinning the spinner twice are listed. We will list an R for red, a B for blue, and a Y for yellow. The sample space is: {BB, BR, BY, RR, RB, RY, YY, YR, YB} Note: All outcomes within the sample space are equally likely. The probability of each individual outcome happening is 1 out of 9 or 1 9. Blue is to appear twice. Therefore, the event will only occur once as shown above. Next, list the probability. The probability can be calculated by using the formula below. P(E) = the number of times blue appears twice on the spinner the number of possible outcomes P(E) = 1 9 Divide to get the or 11%. There are some basic properties that occur when there is a sample space with equally likely outcomes. Basic Properties of Probability Assume that S is a sample space for some experiment and E is an event in S P(E) 1 (The probability of the event is between 0 and 1.) 2. P( ) = 0 (The probability of an empty set is 0.) 3. P(S) = 1 (The sum of the probabilities of all outcomes within a set is 1.) MAT 1301, Liberal Arts Math 4

5 The diagram represents how the probability can be interpreted. This diagram can UNIT be x found STUDY on GUIDE page 646 of the textbook. Intuitive Meaning of Probability (Pirnot, 2014, p. 646) In the previous example, it was stated that the probability of blue appearing twice in two spins of the spinner was 11%. Looking at the chart, this event is not likely to occur. It is important to recall the counting principles learned in the previous unit to help find the sample spaces of some experiments. The next example uses the fundamental counting principle to find the number of outcomes in an experiment. An individual is rolling two four-sided dice. One die has the numbers 1, 2, 3, and 4 on its faces. The other die has the numbers 3, 4, 5, and 6 on its faces. a. How many elements are in the sample space? b. Express the event the total showing equals seven as a set. c. What is the probability that the total showing is equal to seven? a. How many elements are in the sample space? The fundamental counting principle from Section 12.2 of your textbook can be used to count the elements of the sample space: 4 4 = 16. Therefore, there are 16 outcomes in the sample space. b. Express the event the total showing equals seven as a set. First, a table can be created showing all possible outcomes. List the outcomes as ordered pairs in the form (value of first die, value of second die). The highlighted cells represent those outcomes where the total showing equals seven. This event is notated by the set: {(1, 6), (2, 5), (3, 4), (4, 3)} MAT 1301, Liberal Arts Math 5

6 c. What is the probability that the total showing is equal to seven? From (b), we found that a total of seven can occur in 4 ways. From (a), we found that there are 16 possible outcomes. Therefore, the probability of the total showing being seven is A cookie is selected at random from a basket of Girl Scout Cookies containing 16 Thin Mints, 28 Caramel DeLites, 12 Shortbreads, and 24 Shout Outs. Find the probability that the specified event occurs. Event: Neither a Thin Mint nor a Shortbread is selected. First, find the sample space. The sample space consists of the total number of cookies: = 80 cookies Next, find the number of cookies that are neither Thin Mint nor Shortbread. Since there are 28 Caramel DeLites and 24 Shout Outs there are = 52 out of 80 cookies that are neither a Thin Mint nor Shortbread. The probability that neither a Thin Mint nor a Shortbread is selected is P(neither Thin Mint or Shortbread) = 52 = 13 = 0.65 = 65% Probability and Genetics Page 651 of the textbook explains how the Austrian monk Gregor Mendel discovered dominant and recessive genes. Probability theory helped Mendel explain the genetic theory that he discovered. One of the most useful tools used in showing the possible genetic outcomes is called the Punnett square. The Punnett square is a diagram that is used to predict the outcomes of an experiment. The diagram below is an example of a Punnett square. Each cell of a Punnett square is an outcome of the experiment. The first cell, AA, is found by pairing the letter in row 1 with the letter in column 1. The second cell, Aa, is found by pairing the letter in row 1 with the letter in column 2. The same pattern is followed to complete the last two cells. MAT 1301, Liberal Arts Math 6

7 Sickle-cell anemia is a serious inherited disease that is about 30 times more likely to occur in African American babies than in non-african American babies. A person with two sickle-cell genes will have the disease, but a person with only one sickle-cell gene will be a carrier of the disease. If one parent of a child has sickle-cell anemia and the other parent is a carrier of sickle-cell anemia, what is the probability that the child has sickle-cell anemia? Construct a Punnett square to show the probabilities for the offspring. First, construct a Punnett square to show the probabilities. The rows of the Punnett square represent the first parent, and the columns of the Punnett square represent the second parent. The problem states that a person with two sickle-cell genes will have the disease, but a person with only one sickle-cell gene will be a carrier of the disease. Therefore, the first parent has sickle-cell anemia, so both rows are labeled with an s. The second parent is a carrier of sickle cell anemia, so one column is labeled with s and the other column with an n. Denote the sickle-cell gene by s and denote the normal gene by n. To find the probability that the child has the disease, count the number of cells that contain ss and divide by the total number of cells. The probability that the child has the disease is 2 = 1 = = 50%. 4 2 Odds As discussed, probability is the likeliness that something will happen. The term odds is found when calculating the probability against something happening. For example, if you flip a coin and want to find the odds against flipping a tail, you would say that the odds against flipping a tail are 2 to 1. This can be written as a ratio, 2:1, or as a fraction, 2/1. Odds Against an Event If the outcomes of a sample space are equally likely, then the odds against an event E are simply the number of outcomes that are against E compared with the number of outcomes in favor of E occurring. We would write these odds as n(e ):n(e), E is the complement of event E. Complement of Event E The complement is the number of times E does not happen. For example, if E = the event of it raining outside, then E = the event of it not raining outside. The complement is calculated as follows: P(E ) = 1 P(E) Probability Formula for Computing the Odds If E is the complement of the event E, then the odds against E are P(E ) P(E) MAT 1301, Liberal Arts Math 7

8 Event: A total of nine shows when two fair dice are rolled. a. Find the probability of the given event. b. Find the odds against the given event. Start by listing all the possible options for rolling two dice in a table. a. Find the probability of the given event. A total of nine can occur in four ways: {(3,6), (4,5), (5,4), (6,3)} Since there are 36 pairs that can occur, b. Find the odds against the given event. Use the formula, P(E ) P(E) P(E) = 4 36 = % 9 First, find P(E ). (Hint: Use the value for P(E) in part a to calculate P(E ). P(E ) = 1 P(E) = = = 8 9 Since P(E) = 1 and 9 P(E ) = 8, the odds against the event are 9 The answer is 8 to 1 or 8:1. 8 P(E ) P(E) = 9 = In horseracing, a trifecta is a race in which one must pick the first-, second-, and third-place winners in their proper order to win a payoff. If the odds against event E are 5 to 2, what is the probability of E? MAT 1301, Liberal Arts Math 8

9 The odds are 5:2. Therefore, P(E ) P(E) = 5 2. To the probability, work backwards as follows: Thus, the probability of E or P(E) is 2 7. If P(E) = 0.45, then what are the odds against E? The formula for finding the odds against E is Since P(E) = 0.45, P(E ) = 1 P(E) = = Therefore, P(E ) P(E). If the odds against the U.S. men s volleyball team defeating China in the Summer Olympic Games were 9 to 3, what is the probability that the United States defeated China? The odds are 9:3. Therefore, P(E ) P(E) = 9 3. To the probability, work backwards as follows: = 12 = P(E ) 3 P(E) 12 Thus, the probability of defeating China is 3 12 = 1 4 MAT 1301, Liberal Arts Math 9

10 13.2 Complements and Unions of Events: In this section, some mathematical UNIT formulas x STUDY that GUIDE provide a simpler way of applying the principles of probability will be learned. Complements of Events Complements were used in the examples above to compute the odds against an event. Here, how complements relate to the probability and the sample space of an event will be explained. Imagine that the rectangle and oval in the diagram are the sample space of an experiment and that the sum of probabilities in the sample space is one. The rectangle and the oval contain all possible outcomes of the experiment. Therefore, the probability of the outcomes represented by the tan-colored shading can be found by subtracting the probability of the outcomes in the oval from 1. Looking at this diagram, the following basic probability formulas can be identified. One has already been introduced. P(E ) = 1 P(E) P(E) + P(E ) = 1 If the probability that a vaccine one took will prevent him/her from getting the flu is 0.828, what is the probability that someone else will get the flu? Let A be the event that an individual does not get the flu, and let A be the event that the individual does get the flu. The sample space can be illustrated by the diagram: MAT 1301, Liberal Arts Math 10

11 The complement formula will be used to solve. There is a chance of one UNIT not getting x STUDY the GUIDE flu. So, P(A) = To find the probability of getting the flu, P(A ) will be found. P(A ) = 1 P(A) P(A ) = P(A ) = or 17.2% Thus, the probability of one getting the flu is 17.2%. If two dice are rolled, find the probability that the total showing is less than 10 by using the complement formula. First, find out how many possible combinations equal 10 or greater. This can be done by listing all possible combinations in the form of an ordered pair. The first number in the ordered pair represents the digit that appears on die 1 and the second in the order pair represents the digit that appears on die 2. (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (1,4) (2,4) (3,4) (4,4) (5,4) (6,4) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) From this, the total of each set of dice can be added up. As a result, there are six possible combinations, out of 36 total combinations, that are equal to 10 or greater. {(4,6), (5,5), (6,4), (5,6), (6,5), and (6,6)}. Therefore, the probability that the sum is 10 or greater is: 6 36, which reduces to 1 6 Substituting into the complement formula, the probability that the total showing is less than 10 can be found: P(E ) = 1 P(E) P(E ) = P(E ) = P(E ) = 5 6 MAT 1301, Liberal Arts Math 11

12 Unions of Events In probability, the words union and intersection are used to join events. These operations are more specifically used when you want to find the probability of multiple events within the same sample s. Notation: Express the union of two events by the symbol: (E F). Express the intersection of two events by the symbol: (E F). Note: The intersection is the overlapping of outcomes between two events. Rule for Computing the Probability of a Union of Two Events If E and F are events, then P(E F) = P(E) + P(F) P(E F) If E and F have no outcomes in common, they are called mutually exclusive events. In this case, P(E F) = P(E) + P(F) If a single card is drawn from a standard 52-card deck, what is the probability that it is either a face card or a red card? The probability of a union of two events can be used for this example. The different variables of the equation will first need to be identified. There are a total of 12 face cards in a deck, so: P(F) = MAT 1301, Liberal Arts Math 12

13 There are a total of 26 red cards in a deck, so: P(R) = There are a total of 6 face cards that are also red, so: P(F R) = 6 52 Now that the variables have been identified, substitute and solve: P(F R) = P(F) + P(R) P(F R) P(F R) = P(F R) = P(F R) = = 8 13 The next two examples use the rule for computing the probability of two events to find missing information. To solve the problems, first plug in all values given and perform algebraic operations to calculate the missing piece. Assume that A and B are events. If P(A B) = 0.75, P(B) = 0.45, and P(A) = 0.60, find P(A B). Because three out of the four values in the probability of a union of two events formula are given, substitute and solve algebraically as follows: P(A B) = P(A) + P(B) P(A B) 0.75 = P(A B) 0.75 = P(A B) 0.75 = 1.05 P(A B) = P(A B) 0.30 = P(A B) P(A B) = 0.30 Assume that A and B are events. If P(A B) = 0.60, P(B) = 0.45, and P(A B) = 0.20, find P(A). MAT 1301, Liberal Arts Math 13

14 Because three out of the four values in the probability of a union of two events formula are given, substitute and solve algebraically as follows: P(A B) = P(A) + P(B) P(A B) 0.60 = P(A) = P(A) = P(A) P(A) = Conditional Probability and Intersections of Events: Assume that you have the choice between eating four desserts. The probability that you will choose any one dessert is 25%. Suppose that your friend ate one of the desserts. Now, you only have the option of choosing three desserts, so the probability of choosing any one dessert is 33%. Sometimes the occurrence of one event can affect the probability of a second event. This concept is called conditional probability and is found when one knows that something else has already happened. Conditional Probability When the probability of event F is determined, assuming that the event E has already occurred, this is called the conditional probability of F given E. We denote this probability as P(F E). P(F E) is read as the probability of F given that E has occurred or the probability of F given E. Special Rule for Computing P(F E) by Counting If E and F are events in a sample space with equally likely outcomes, then P(F E) = n(e F) n(e) Assume that you roll two dice and the total showing is odd. What is the probability that the total is seven? First, the number of all possible outcomes in the form of an ordered pair needs to be found. The first number in the ordered pair represents the digit that appears on die 1, and the second number in the ordered pair represents the digit that appears on die 2. (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (1,4) (2,4) (3,4) (4,4) (5,4) (6,4) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) From this, we can see that there are 36 possible outcomes. F will be the event of rolling an odd total, and E will be the event of rolling a total of five. Therefore, F={((2,1), (4,1), (6,1), (1,2), (3,2), (5,2), (2,3), (4,3), (6,3), (1,4), (3,4), (5,4), (2,5), (4,5), (6,5), (1,6), (3,6), (5,6)}. MAT 1301, Liberal Arts Math 14

15 Set E consists of all pairs in F that total seven. E = {(6,1), (5,2), (4,3), (3,4), (2,5), (1,6)} Therefore, the probability of F given E is the number of outcomes in E divided by the number of outcomes in F. P(F E) = P(F E) = n(e F) n(e) 6 18 = % The probability for computing P(F E) can also be found by using the probability of the intersection instead of the number of outcomes in the intersection as used above. This is the general rule for computing P(F E). General Rule for Computing P(F E) If E and F are events in a sample space, then P(F E) = P(E F) P(E) Assume that two fair dice are being rolled. First, compute P(F) and then P(F E). Explain why one would expect the probability of F to change as it did when the condition that E had occurred was added. E an even total shows on the dice. F the total is four. If all of the combinations that can be rolled with two dice were listed, we would see that there are three out of thirty-six different combinations that will add up to a total of four. These would be: Therefore, it can be seen that: {(1,3) (3,1) (2,2)} P(F) = 3 36 = 1 12 If all of the combinations that have an even total were listed, it would be seen that there are eighteen out of thirty-six combinations. These would be: {(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6)}. Therefore, it can be seen that: P(E) = = 1 2 Now, find P(E F). It can be seen that {(1,3) (3,1) and (2,2)} are the only three totals that are present in both groups, therefore: P(E F) = 3 36 MAT 1301, Liberal Arts Math 15

16 Now, substitute into the final equation, solve, and simplify: P(F E) = 3 36 = = 6 36 = Draw a single card from a standard 52-card deck. Find the probability: First find the following: P(even numbered card nonface card) P(nonface card) and P(nonface card even numbered card) There are a total of 40 non-faced cards in a deck, therefore: Of those 40 non-faced cards, half are even. Therefore: This results in: The Intersection of Events P(nonface card) = P(nonface card even numbered card) = P(even numbered card nonface card) = P(even numbered card nonface card) = P(nonface card even numbered card) P(nonface card) = = The probability of an intersection of two events can be found by using the formula for conditional probability. Rule for Computing the Probability of the Intersection of Events If E and F are two events, then P(E F) = P(E) P(F E). Assume that two cards are drawn from a standard 52-card deck. Find the probability that two jacks are drawn. Assume that the cards are drawn without replacement. There are 52 cards in a deck, and four are jacks. A jack is to be drawn on the first try. Therefore, P(jack) = 4 52 = 1 13 MAT 1301, Liberal Arts Math 16

17 Now, another jack is to be drawn on the second try. Remember that the cards UNIT are not x STUDY being replaced, GUIDE so there are only 51 total cards left to choose from, and only three of them are jacks. The probability of drawing another jack is P(another jack jack) = 3 51 Therefore, the probability drawing two jacks is represented by P(jack another jack) = P(jack) P(another jack jack) = = Probability Trees = 1 =.0045 = 0. 45% 221 Tree diagrams are beneficial when computing the probability of an event. These diagrams are called probability trees and help to visualize the computations that must be performed to get the resulting probability. Once a probability tree is constructed, it can be seen that each branch of the tree represents the outcomes of the experiment. The resulting probability can be found by multiplying the probabilities found along the branch representing that outcome. An individual wants to purchase a DVD drive for his/her laptop computer. Assume that 65% of the drives are made outside the United States. Of the U.S.-made drives, 4% are defective; of the foreign-made drives, 6% are defective. Determine the probability that the drive an individual purchases is foreign-made and is defective. Round to three decimal places. This problem can be solved by constructing a probability tree diagram. First, draw two branches. The first branch represents U.S.-made drives and the second branch represents foreign-made drives. The problem states that 65% of the drives were made outside the United States. Therefore, the foreign branch will be labeled with a probability of 0.65, and the U.S. branch will be labeled with the probability of = This is shown above. MAT 1301, Liberal Arts Math 17

18 Next, two branches will be added to the foreign branch and to the U.S. branch UNIT to identify x STUDY the probability GUIDE of a drive being defective or non-defective. The problem states that of the U.S.-made drives, 4% are defective. Therefore, the defective branch on the U. S. side will be labeled with a probability of The non-defective branch will be labeled with a probability of 0.96 because = The problem also states that of the foreign-made drives, 6% are defective. Therefore, the defective branch on the foreign side will be labeled with a probability of The non-defective branch will be labeled with a probability of 0.94 because = This is shown in the diagram above. Next, multiply the probability of each possible outcome. The probability the drive purchased is foreign-made and is defective is or 3.9%. Dependent and Independent Events Conditional probability has taught that sometimes the outcome of a previous event can affect the next event. This happens when events are dependent on each other. If two events are independent of each other, then their probability will not be affected. The definitions of dependent and independent events will be explained to better understand when certain probability principles should be used. MAT 1301, Liberal Arts Math 18

19 Independent Events Events E and F are independent events if P(F E) = P(F). In other words, the occurrence of event E did not influence event F. Dependent Events Events E and F are dependent events if P(F E) P(F). In other words, the occurrence of event E will affect the probability of event F. Two balls are selected randomly without replacement from the urn below. Is the event the first ball is red and the second ball is green independent or dependent? Drawing balls from an urn (Pirnot, 2014, p. 673) This event is dependent because the balls are not replaced after each event. Therefore, the event of drawing a second ball is dependent on the event of drawing the first ball. For more information regarding probability and odds as well as a review of information covered in the last unit, please view the following interactive lesson: /redirect_index_63.html Reference Pirnot, T. L. (2014). Mathematics all around (5th ed.). Boston, MA: Pearson. MAT 1301, Liberal Arts Math 19

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