Chapter 7: Section 7-1 Probability Theory and Counting Principles D. S. Malik Creighton University, Omaha, NE D. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 1 / 20
The theory of probability plays a crucial role in making inferences. Intuitively, probability measures how likely something is to occur. According to Pierre Simon de Laplace probability is the ratio of the number of favorable cases to the total number of cases, assuming that all of the various cases are equally possible. To apply this definition, one needs to know the number of favorable cases and the total number of cases, so we need to know various counting techniques. In the previous chapter, we assumed that all of the outcomes of an experiment are equally likely. However, in general this may not be true. Suppose that a weighted coin is tossed and the chances of getting a head are 65%. Then the chances of getting a tail are 35%. In this case, we could assign Pr[H] = 0.65 and Pr[T ] = 0.35. Let us note the following: 1. S = {H, T }. 2. Pr[H] = 0.65 and Pr[T ] = 0.35. 3. Pr[H] + Pr[T ] = 0.65 + 0.35 = 1. D. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 2 / 20
Definition Let S = {s 1, s 2,..., s k } be a sample space with k outcomes. A probability assignment to the outcome s i of S is an assignment of a real number p i, i = 1, 2,..., k, such that the following holds: (i) 0 p i 1, i = 1, 2,..., k. (ii) p 1 + p 2 + + p k = 1. If p i is the probability of the outcome s i, then we write Pr[s i ] = p i, i = 1, 2,..., k. Example Suppose that a coin is tossed. Then S = {H, T }. Let Pr[H] = 0.53 and Pr[T ] = 0.47.. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 3 / 20
Example Suppose that an experiment has 5 outcomes, S = {s 1, s 2, s 3, s 4, s 5 }. Suppose that the probability of the outcomes are given by the following table. Outcome s 1 s 2 s 3 s 4 s 5 Probability 0.05 0.35 0.10 0.30 0.20 Now 0 Pr[s i ] 1 for all i and Pr[s 1 ] + Pr[s 2 ] + Pr[s 3 ] + Pr[s 4 ] + Pr[s 5 ] = 0.05 + 0.35 + 010 + 0.30 + 0.20 = 1.0. Thus, we have a probability assignment.. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 4 / 20
Example Suppose that a weighted die is rolled such that an even number is twice as likely to be rolled as an odd number. All even numbers are equally likely to occur, and all odd numbers are equally likely to occur. Let us determine a probability assignment for this experiment. Suppose that Pr[1] = p. Then Now, Pr[1] = Pr[3] = Pr[5] = p and Pr[2] = Pr[4] = Pr[6] = 2p. This implies that Pr[1] + Pr[2] + Pr[3] + Pr[4] + Pr[5] + Pr[6] = 1 p + 2p + p + 2p + p + 2p = 1 9p = 1 p = 1 9. Pr[1] = Pr[3] = Pr[5] = 1 9 and Pr[2] = Pr[4] = Pr[6] = 2 9.. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 5 / 20
Definition Let S = {s 1, s 2,..., s k } be a sample space of an experiment such that Pr[s i ] = p i, i = 1, 2,..., k, is a probability assignment. Let E = {y 1,..., y t } be an event is S. The probability of the event E, written Pr[E ], is defined by Moreover, Pr[ ] = 0. Pr[E ] = Pr[y 1 ] + + Pr[y t ]. Example Suppose that a die is rolled such that Pr[1] = 0.10, Pr[2] = 0.15, Pr[3] = 0.20, Pr[4] = 0.10, Pr[5] = 0.25, Pr[6] = 0.20. Note that this is probability assignment. Let E = {1, 3, 6}. Then Pr[E ] = Pr[1] + Pr[3] + Pr[6] = 0.10 + 0.20 + 0.20 = 0.50. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 6 / 20
Theorem Let S = {s 1, s 2,..., s k } be a sample space with k outcomes. Suppose that Pr[s i ] = p i, i = 1, 2,..., k, is a probability assignment for the experiment. Let E and F be events of the experiment. Then (i) Pr[ ] = 0. (ii) 0 Pr[E ] 1. (iii) Pr[S] = 1. (iv) Pr[E ] = 1 Pr[E ]. (v) If E F =, then Pr(E F ) = Pr[E ] + Pr[F ]. (vi) Pr(E F ) = Pr[E ] + Pr[F ] Pr[E F ].. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 7 / 20
Example A card is drawn from a well-shuffl ed deck of 52 cards. (a) Let E be the event that the drawn card is a king. Then n(e ) = 4. So Pr[E ] = n(e ) n(s) = 4 52 = 1 13. (b) Suppose that we want to find the probability that the drawn card is not a king. Let E be the event that the drawn card is a king. Then Pr[E ] = 1 13. Note that E is the event that the drawn card is not a king. Now Pr[E ] = 1 Pr[E ] = 1 1 13 = 12 13. Hence, the probability that the drawn card is not a king is 12 13. Note that you can also answer this question by observing that there are 48 cards that are not kings. Hence, the probability that the drawn card is not a king is 48 52 = 12 13.. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 8 / 20
Mutually Exclusive Events Definition Let S be a sample space of an experiment and E and F are events in S. Then E and F are called mutually exclusive if E F =. Theorem Let S be a sample space of an experiment and E and F are mutually exclusive events in S. Then Pr[E F ] = 0.. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 9 / 20
Example A card is drawn at random from a well-shuffl ed deck of 52 cards. What is the probability that the drawn card is a face card and an ace? Let E be the event that the drawn card is a face card. Then E is the set of kings, queens, and jacks. Let F be the event that the drawn card is an ace. Then F is the set of all aces. It follows easily that E F =. Thus, the events E and F are mutually exclusive. Moreover, Pr[E F ] = 0. Hence, the probability that the drawn card is a face card and an ace is 0.. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 10 / 20
Odds in Favor of an Event Definition Let E be an event such that Pr[E ] = 0. Then the odds in favor of the event E are defined to be Pr[E ] Pr[E ]. Remark If odds in favor of an event E are m n, then sometimes we write this as m to n or m : n.. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 11 / 20
Example The probability of winning the next basketball game is 0.48. Let E be the event that the next basketball game will be won. Then This implies that Thus, the odds in favor of E are Pr[E ] = 0.48. Pr[E ] = 1 Pr[E ] = 1 0.48 = 0.52. Pr[E ] Pr[E ] = 0.48 0.52 = 48 52 = 12 13. Hence, the odds in favor of winning the next basketball game are 12 to 13.. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 12 / 20
Theorem Let E be an event such that the odds in favor of E are m to n, i.e., Pr[E ] Pr[E ] = m, n = 0. n Then and Pr[E ] = Pr[E ] = m m + n n m + n.. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 13 / 20
Example The odds in favor of winning a bingo game are 5 to 17. Let E be the event of wining the bingo game. Then Here m = 5 and n = 17. Thus, Pr[E ] Pr[E ] = 5 17. Pr[E ] = 5 5 + 17 = 5 22. So the probability of winning the bingo game is 5 22. From this we can also conclude that the probability of losing the bingo game is 1 5 22 = 17 22. Note that by the previous theorem, Pr[E ] = 17 5+17 = 17 22.. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 14 / 20
Empirical Probabilities When a fair coin is tossed, we know the exact probability of getting a head or a tail. No matter how many times we toss a coin, the probability of getting a head is always the same. There are situations when the exact probabilities cannot be assigned. For example, when a new drug is introduced in a market, it is not possible to exactly know how many people will get severe side effects. In cases such as these, the probabilities are assigned based on observations. Suppose that in the clinical trial of the drug, 100 people participated and 2% had severe side effects. Based on this observation, we can say that the probability that a person will get a severe side effect is 0.02. Such probabilities are called empirical probabilities. D. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 15 / 20
Example At the May graduation, the numbers of students graduating with specific majors are given in the following table. Major No of Graduates Major No of Graduates Accounting 275 History 80 Biology 176 Mathematics 35 Chemistry 190 Physics 25 Communications 75 Others 475 English 55 The total number of graduates is 275 + 176 + 190 + 75 + 55 + 80 + 35 + 25 + 475 = 1386. We now use the given data to assign probabilities as follows. The number of students graduating with an accounting major is 275. So Pr[Accounting Major] = 275 1386 0.1984.. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 16 / 20
Exercise: Let S = {s 1, s 2, s 3, s 4 } be a sample space. Determine whether the following is a probability assignment on S. s s 1 s 2 s 3 s 4 p 0.25 0.20 0.35 0.20 Solution: From the table, 0 Pr[s i ] 1 and Pr[s 1 ] + Pr[s 2 ] + Pr[s 3 ] + Pr[s 4 ] = 0.25 + 0.20 + 0.35 + 0.20 = 1. Hence, we have a probability assignment. D. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 17 / 20
Exercise: Let S = {s 1, s 2, s 3, s 4, s 5 } be a sample space. Determine whether the following is a probability assignment on S. s s 1 s 2 s 3 s 4 s 5 p 0.15 0.40 0.25 0.20 0.5 Solution: From the table, 0 Pr[s i ] 1. However, Pr[s 1 ] + Pr[s 2 ] + Pr[s 3 ] + Pr[s 4 ] + Pr[s 5 ] = 0.15 + 0.40 + 0.25 + 0.20 + 0.05 = 1.05 > 1 This implies that the given table does not give a probability assignment on S. D. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 18 / 20
Exercise: A survey revealed that 45% of people get their news by reading the newspaper, 65% get their news by watching TV, and 30% get their news by both the newspaper and the TV. What is the probability that a randomly selected person gets his/her news by either the newspaper or TV? Solution: Let N be the set of people who get their news by reading the newspaper and T be the set of people who get their news by watching TV. Then Now Pr[N] = 0.45, Pr[T ] = 0.65, and Pr[N T ] = 0.30. Pr[N T ] = Pr[N] + Pr[T ] Pr[N T ] = 0.45 + 0.65 0.30 = This implies that the probability that a randomly selected person gets his/her news by either the newspaper or TV is 0.80. D. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 19 / 20
Exercise: A card is drawn from a well-shuffl ed deck of 52 cards. Find the odds in favor of drawing a queen or a black card. Solution: Let E be the event that the drawn card is a queen or a black card. There are 26 black cards, 4 queens, and 2 of the queens are black. Then n(e ) = 26 + 4 2 = 28. So This implies that Pr[E ] = 28 52. Thus, Pr[E ] = 1 Pr[E ] = 1 28 52 = 24 52. 28 Pr[E ] Pr[E ] = 52 = 7 6. Hence, the odds in favor of drawing a queen or a black card is 7 to 6. D. S. Malik Creighton University, Omaha, NE Chapter () 7: Section 7-1 Probability Theory and Counting Principles 20 / 20 24 52