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4 th EDITION

College Algebra & Trigonometry and Precalculus. 4 th EDITION. 11.7. Basics of Probability. Basic Concepts Complements and Venn Diagrams Odds Union of Two Events Binomial Probability. Basic Concepts.

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4 th EDITION

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  1. College Algebra & Trigonometry and Precalculus 4th EDITION

  2. 11.7 Basics of Probability Basic Concepts Complements and Venn Diagrams Odds Union of Two Events Binomial Probability

  3. Basic Concepts Consider an experiment that has one or more possible outcomes, each of which is equally likely to occur. For example, the experiment of tossing a fair coin has 2 equally likely possible outcomes: landing heads up (H) or landing tails up (T). Also, the experiment of rolling a fair die has 6 equally likely outcomes: landing so the face that is up shows 1, 2, 3, 4, 5, or 6 dots.

  4. Basic Concepts The set S of all possible outcomes of a given experiment is called the sample space for the experiment. (In this text, all sample spaces are finite.) One sample space for the experiment of tossing a coin could consist of the outcomes H and T. This sample space can be written as Use set notation. Similarly, a sample space for the experiment of rolling a single die once is

  5. Basic Concepts Any subset of the sample space is called an event. In the experiment with the die, for example, “the number showing is a 3” is an event, say E1, such that E1 = {3}. “The number showing is greater than 3” is also an event, say E2,such that E2 = {4,5,6}. To represent the number of outcomes that belong to event E, the notation n(E) is used. Then n(E1) = 1 and n(E2) = 3.

  6. Basic Concepts The notation P(E) is used for the probability of an event E. If the outcomes in the sample space for an experiment are equally likely, then the probability of event E occurring is described here next.

  7. Probability of Event E In a sample space with equally likely outcomes, the probability of an event E, written P(E), is the ratio of the number of outcomes in sample space S that belong to event E, n(E), to the total number of outcomes in sample space S, n(S). That is,

  8. Basic Concepts To use this definition to find the probability of the event E1 in the die experiment, start with the sample space, S = {1, 2, 3, 4, 5, 6} and the desired event, E1 = {3}. Since n(E1) = 1 and n(S) = 6,

  9. FINDING PROBABILITIES OF EVENTS Example 1 A single die is rolled. Write each event in set notation and give the probability of the event. (a) E3: the number showing is even Solution (a) Since E3 = {2, 4, 6}, n(E3) = 3. As given earlier, n(S) = 6, so

  10. FINDING PROBABILITIES OF EVENTS Example 1 A single die is rolled. Write each event in set notation and give the probability of the event. (b) E4: the number showing is greater than 4 Solution (b) Again n(S) = 6. Event E4 = {5, 6}, with n(E4) = 2.

  11. FINDING PROBABILITIES OF EVENTS Example 1 A single die is rolled. Write each event in set notation and give the probability of the event. (c) E5 : the number showing is less than 7 Solution (c) E5 = {1, 2, 3, 4, 5, 6}, and P(E5) = This event is certain to occur every time the experiment is performed. An event that is certain to occur always has probability 1.

  12. FINDING PROBABILITIES OF EVENTS Example 1 A single die is rolled. Write each event in set notation and give the probability of the event. (d) E6 : the number showing is 7 Solution (d) The probability of an impossible event, such as E6 is always 0, since none of the outcomes in the sample space satisfy the event. For any event E, P(E) is between 0 and 1 inclusive.

  13. Complements and Venn Diagrams The set of all outcomes in the sample space that do not belong to event E is called the complement of E, written E'.For example, in the experiment of drawing a single card from a standard deck of 52 cards, let E be the event “the card is an ace.” Then E' is the event “the card is not an ace.” From the definition of E', for an event E,

  14. Note A standard deck of 52 cards has four suits: hearts , diamonds , spades , and clubs , with thirteen cards of each suit. Each suit has a jack, a queen, and a king (sometimes called the “face cards”), an ace, and cards numbered from 2 to 10. The hearts and diamonds are red and the spades and clubs are black. We will refer to this standard deck of cards in this section.

  15. Probability Concepts Probability concepts can be illustrated using Venn diagrams, as shown in the diagram below. The rectangle shown here represents the sample space in an experiment. The area inside the circle represents event E, while the area inside the rectangle, but outside the circle, represents event E'.

  16. USING THE COMPLEMENT Example 2 In the experiment of drawing a card from a well-shuffled deck, find the probabilities of event E, the card is an ace, and of event E'. Solution Since there are 4 aces in the deck of 52 cards, n(E) = 4 and n(S) = 52. Therefore,

  17. USING THE COMPLEMENT Example 2 Of the 52 cards, 48 are not aces, so In Example 2, This is always true for any event E and its complement E'. That is,

  18. This can be restated as or These two equations suggest an alternative way to compute the probability of an event. For example, if it is known that then

  19. Odds Sometimes probability statements are expressed in terms of odds, a comparison of P(E) with P(E'). The odds in favor of an event E are expressed as the ratio of P(E) to P(E‘)or as the quotient For example, if the probability of rain can be established as ⅓, the odds that it will rain are

  20. Odds On the other hand, the odds that it will not rain are 2 to 1 (or ⅔ to ⅓). If the odds in favor of an event are, say, 3 to 5, then the probability of the event is ⅜ and the probability of the complement of the event is ⅝. If the odds favoring event E are m to n, then

  21. FINDING ODDS IN FAVOR OF AN EVENT Example 3 A shirt is selected at random from a dark closet containing 6 blue shirts and 4 shirts that are not blue. Find the odds in favor of a blue shirt being selected. Solution Let E represent “a blue shirt is selected.” Then, Therefore, the odds in favor of a blue shirt being selected are

  22. Union of Two Events Since events are sets, we can use set operations to find the union of two events. Suppose a fair die is rolled. Let H be the event “the result is a 3,” and K the event “the result is an even number.” From the results earlier in this section,

  23. Union of Two Events Notice that P(H) + P(K) = P(HK). Before assuming that this relationship is true in general, consider another event G for this experiment, “the result is a 2.”

  24. Union of Two Events In this case, P(G) + P(K) P(GK). As shown here, the graphic suggests the difference in the two examples previously shown comes from the fact that events H and K cannot occur simultaneously. Such events are called mutually exclusive events.

  25. Union of Two Events In fact, H  K = , which is always true for mutually exclusive events. Events G and K, however, can occur simultaneously. Both are satisfied if the result of the roll is a 2, the element in their intersection This example suggests the property shown on the next slide.

  26. Probability of the Union of Two Events For any events E and F,

  27. FINDING PROBABILITIES OF UNIONS Example 4 One card is drawn from a well-shuffled deck of 52 cards. What is the probability of the following outcomes? (a) The card is an ace or a spade. Solution (a) The events “drawing an ace” and “drawing a spade” are not mutually exclusive since it is possible to draw the ace of spades, an outcome satisfying both events. The probability is P(ace or spade) = P(ace) + P(spade) – P (ace and spade)

  28. FINDING PROBABILITIES OF UNIONS Example 4 One card is drawn from a well-shuffled deck of 52 cards. What is the probability of the following outcomes? (b) The card is a 3 or a king. Solution (b) “Drawing a 3” and “drawing a king” are mutually exclusive events because it is impossible to draw one card that is both a 3 and a king.

  29. FINDING THE PROBABILITIES OF UNIONS Example 5 Suppose two fair dice are rolled. Find each probability. (a) The first die shows a 2, or the sum of the two dice is 6 or 7. Solution (a) Think of the two dice as being distinguishable, one red and one green for example. (Actually, the sample space is the same even if they are not apparently distinguishable.)

  30. FINDING THE PROBABILITIES OF UNIONS Example 5 A sample space with equally likely outcomes is shown here, where (1, 1) represents the event “the first die (red) shows a 1 and the second die (green) shows a 1,” (1, 2) represents “the first die shows a 1 and the second die shows a 2,” and so on

  31. FINDING THE PROBABILITIES OF UNIONS Example 5 Let A represent the event “the first die shows a 2,” and B represent the event “the sum of the two dice is 6 or 7.” Event A has 6 elements, event B has 11 elements, and the sample space has 36 elements. Thus, so

  32. FINDING THE PROBABILITIES OF UNIONS Example 5 Suppose two fair dice are rolled. Find each probability. (b) The sum of the dots showing is at most 4. Solution (b) “At most 4” can be written as “2 or 3 or 4.” (A sum of 1 is meaningless here.) Since the events represented by “2,” “3,” or “4” are mutually exclusive, P (at most 4) = P(2 or 3 or 4) = P(2) + P(3) + P(4). (*)

  33. FINDING THE PROBABILITIES OF UNIONS Example 5 The sample space for this experiment includes the 36 possible pairs of numbers shown here. The pair (1,1) is the only one with a sum of 2, so

  34. FINDING THE PROBABILITIES OF UNIONS Example 5 Also since both (1,2) and (2,1) give a sum of 3. The pairs (1,3), (2,2), and (3,1) have a sum of 4, so

  35. FINDING THE PROBABILITIES OF UNIONS Example 5 Substituting into equation (*) shown here again gives P (at most 4) = P(2 or 3 or 4) = P(2) + P(3) + P(4).

  36. Properties of Probability For any events E and F: • 1. 0 ≤ P(E) ≤ 1 • P(an impossible event) = 0 • 5. P(E or F) = P(EF) = P(E) + P(F) – P(EF). 2. P(a certain event) = 1 4. P(E') = 1 – P(E)

  37. CautionWhen finding the probability of a union, remember to subtract the probability of the intersection from the sum of the probabilities of the individual events.

  38. Binomial Probability A binomial experiment is an experiment that consists of repeated independent trials with only two outcomes in each trial, success or failure. Let the probability of success in one trial be p. Then the probability of failure is 1 – p, and the probability of exactly r successes in n trials is given by

  39. Binomial Probability This expression is equivalent to the general term of the binomial expansion given earlier. Thus the terms of the binomial expansion give the probabilities of exactly r successes in n trials, for 0 ≤ r ≤ n, in a binomial experiment.

  40. FINDING PROBABILITIES IN A BINOMIAL EXPERIMENT Example 6 An experiment consists of rolling a die 10 times. Find each probability. (a) The probability that in exactly 4 of the rolls, the result is a 3 Solution • The probability p of a 3 on one roll is • Here n = 10 and r = 4, so the required probability is

  41. FINDING PROBABILITIES IN A BINOMIAL EXPERIMENT Example 6 An experiment consists of rolling a die 10 times. Find each probability. (b) The probability that in exactly 9 of the rolls, the result is not a 3 Solution (b) The probability is Use n = 10, r = 9, and

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