7.3 Introduction to Probability

7.3 Introduction to Probability A great many problems that come up in applications of mathematics involve random phenomena - those for which exact prediction is impossible. The best we can do is determine the probability of the possible outcomes. In this section, we will formally define chance or probability. We will start with the basic definitions. Definition 1. In probability, an experiment is an activity or occurrence with an observable result. Each repetition of an experiment is called a trial. The possible results of each trial are called outcomes. The set of all possible outcomes for an experiment is the sample space for that experiment. For example, an experiment may comprise of (repeatedly) tossing a coin. Each toss of the coin is a trial. The possible outcomes are heads (h) and tails (t). If S represents the sample space, then S = {h, t}. Example 1. Give the sample space for each experiment. (a) A spinner like the one in the following figure is spun. Solution. The three possible outcomes are 1, 2 and 3, so the sample space is {1, 2, 3} (b) For the purpose of a public opinion poll, respondents are classified as young, middleaged, or senior, and as male or female. Solution. A sample space for this poll could be written as a set of pairs: {(young, male), (young, female), (middle-aged, male), (middle-aged, female), (senior, male), (senior, female)}. Fall 2010 Page 1 Penn State University

(c) An experiment consists of studying the number of boys and girls in families with exactly 3 children. Let b represent boy and g represent girl. Solution. A three-children family can have 3 boys, written bbb, 3 girls, ggg, or various combinations, such as bgg. A sample space with four outcomes (not equally likely) is S 1 = {3 boys, 2 boys and 1 girl, 1 boy and 2 girls, 3 girls}. Note that a family with 3 boys or 3 girls can occur in just one way, but a family of 2 boys and 1 girl or 1 boy and 2 girls can occur in more than one way. If the order of the births is considered, so that bgg is different from gbg or ggb, for example, another sample space is S 2 = {bbb, bbg, bgb, gbb, bgg, gbg, ggb, ggg}. The second sample space, S 2, has equally likely outcomes if we assume that boys and girls are equally likely (we assume this approximately true fact). The outcomes in S 1 are not equally likely, since there is more than one way to get a family with 2 boys and 1 girl (bbg, bgb, or gbb) or a family with 2 girls and 1 boy (ggb, gbg, or bgg), but only one way to get 3 boys (bbb) or 3 girls (ggg). Remark. Example 1(c) shows that an experiment can have more than one sample space. The most convenient sample spaces have equally likely outcomes, but it is not always possible to choose such a sample space. Definition 2. An event is a subset of a sample space. For example, if the sample space for tossing a coin is S = {h, t}, then one event is E = {h}, which represents the outcome heads. Example 2. An ordinary die is a cube whose six different faces show the following numbers of dots: 1, 2, 3, 4, 5, and 6. If the die is fair (not loaded to favor certain faces over others), then any one of the faces is equally likely to come up when the die is rolled. The sample space for the experiment of rolling a single fair die is S = {1, 2, 3, 4, 5, 6}. Some possible events are listed below. The die shows an even number: E 1 = {2, 4, 6}. The die shows a 1: E 2 = {1}. The die shows a number less than 5: E 3 = {1, 2, 3, 4}. The die shows a multiple of 3: E 4 = {3, 6}. Note that n(s) = 6, n(e 1 ) = 3, n(e 2 ) = 1, n(e 3 ) = 4, and n(e 4 ) = 2. observations will be useful later in this section. This type of Example 3. For the sample space S 2 in Example 1(c), write the following events. (a) Event H: the family has exactly two girls Fall 2010 Page 2 Penn State University

Solution. Families with three children can have exactly two girls with either bgg, gbg, or gbg, so event H is H = {bgg, gbg, ggb}. (b) Event K: the three children are the same sex Solution. Two outcomes satisfy this condition: all boys or all girls. So (c) Event J: the family has three girls K = {bbb, ggg}. Solution. Only ggg satisfies this condition, so J = {ggg}. Definition 3. An event which has only one possible outcome is called a simple event. For example, the event J in Example 3(c) is a simple event since it contains only one possible outcome namely, ggg. Definition 4. If an event E equals the sample space, then E is called a certain event. Definition 5. An event E =, then E is called an impossible event, or a null event. Example 4. Suppose a coin is flipped until both a head and a tail appear, or until the coin has been flipped four times, whichever comes first. Write each of the following events in set notation. (a) The coin is flipped exactly three times. Solution. This means that the first two flips of the coin did not include both a head and a tail, so they must both be heads or both be tails. Because the third flip is the last one, it must show the side of the coin not seen on the first two flips. Thus the event is {hht, tth}. (b) The coin is flipped at least three times. Solution. This means that the coin is flipped exactly three times or four times. So the outcomes listed in part (a) are also included in this event. In addition, there is also the possibility that the coin is flipped four times, which only happens when the first three flips are all heads or all tails. Thus the event is (c) The coin is flipped at least two times. {hht, tth, hhhh, hhht, tttt, ttth}. Fall 2010 Page 3 Penn State University

Solution. This event consists of the entire sample space (why?): This is an example of a certain event. (d) The coin is flipped fewer than two times. S = {ht, th, hht, tth, hhhh, hhht, tttt, ttth}. Solution. The coin cannot be flipped fewer than two times under the rules of the experiment, so the event is the empty set. This is an example of an impossible event. Since events are sets, we can use set operations to find unions, intersections, and complements of events. A summary of the set operations for events is given below. Set Operations for Events Let E and F be events for a sample space S. E F occurs when both E and F occur; E F occurs when E or F or both occur; E occurs when E does not occur. Example 5. A study of worker earning the minimum wage grouped such workers into various categories, which can be interpreted as events when a worker is selected at random. Consider the following events: E : worker is under 20; F : worker is female. Describe the following events in words. (a) E Solution. E is the event that the worker is not under 20, that is, the worker is 20 or over. (b) E F Solution. E F is the event that worker is under 20 and not a female, that is, the worker is a male under 20. (c) E F Solution. E F is the event that the worker is under 20 or is a female. Note that this event includes all workers under 20, both male or female and all female workers of any age. Definition 6. Two events, E and F are said to be mutually exclusive events if they cannot occur at the same time that is, E F =. Mutually Exclusive Events Events E and F are mutually exclusive events if E F =. Fall 2010 Page 4 Penn State University

For example, rolling an even number and rolling an odd number are mutually exclusive events in the experiment of rolling a single fair die. Remark. Let E be an event in an experiment. Then E and its complement E are mutually exclusive. By definition, mutually exclusive events are disjoint sets. Example 6. Let S = {1, 2, 3, 4, 5, 6}, the sample space for rolling a single fair die. Let E = {4, 5, 6}, and let G = {1, 2}. Then E and G are mutually exclusive events since they have no outcomes in common: E G =. Definition 7. For sample spaces with equally likely outcomes, the probability of an event is defined as follows. Basic Probability Principle Let S be a sample space of equally likely outcomes, and let event E be a subset of S. Then the probability that event E occurs is P (E) = n(e) n(s). Remark. By this definition, the probability of an event is a number that indicates the relative likelihood of the event. Example 7. Suppose a single fair die is rolled. Use the sample space S = {1, 2, 3, 4, 5, 6} and give the probability of each event. (a) E: the die shows an even number Solution. Here, E = {2, 4, 6}, a set with three elements. Since S contains six elements, (b) F : the die shows a number less than 10 P (E) = 3 6 = 1 2. Solution. Event F is a certain event, with F = {1, 2, 3, 4, 5, 6}, so that P (F ) = 6 6 = 1. Fall 2010 Page 5 Penn State University

(c) G: the die shows an 8 Solution. This event is impossible that is, G =. So P (G) = 0. A standard deck of 52 cards has four suits: hearts ( ), clubs ( ), diamonds ( ), and spades ( ) with 13 cards in each suit. The hearts and diamonds are red, and the spades and clubs are black. Each suit has an ace (A), a king (K), a queen (Q), a jack (J), and cards numbered from 2 to 10. The jack, queen and king are called face cards. Example 8. If a single card is drawn at random from a standard 52-card deck, find the probability of each event. (a) Drawing an ace Solution. There are 4 aces in the deck. The event of drawing an ace is {heart ace, diamond ace, club ace, spade ace}. The sample space contains 52 elements. Therefore, (b) Drawing a face card P (ace) = 4 52 = 1 13. Solution. There are 3 face cards in each of the 4 suits. So there are 12 face cards in the deck. Thus P (face card) = 12 52 = 3 13. (c) Drawing a spade Solution. The deck contains 13 spades, so (d) Drawing a spade or a heart P (spade) = 13 52 = 1 4. Solution. Besides the 13 spades, the deck contains 13 hearts. Moreover, there is no card that is a spade and a heart at the same time. So P (spade or heart) = 26 52 = 1 2. In the preceding examples, the probability of each event was a number between 0 and 1. This is true in general. Any event E is a subset of the sample space S, so 0 n(e) n(s). Since P (E) = n(e) n(s), it follows that 0 P (E) 1. Fall 2010 Page 6 Penn State University

For any event E, 0 P (E) 1. Example 9. The following table gives the number of years of service of senators in the 109th Congress of the United States of America, which convened on January 4, 2005. Years of Service Number of Senators 0 9 50 10 19 28 20 29 15 30 39 4 40 or more 3 Find the probability that a randomly chosen senator of the 109th Congress served 20 29 years when Congress convened. Solution. the probability is found by dividing the number of senators who served 20 29 years by the total number of senators. The total number of senators is 100. Thus, P (20 29 years) = 15 100 = 0.15. Fall 2010 Page 7 Penn State University