Probability, Part 1 We’ve looked at several counting techniques to determine the number of possible arrangements or groupings in a given scenario. These techniques include the multiplication principle, the addition principle, permutations, and combinations.

Probability, Part 1 Together these techniques provide us with a powerful set of tools for calculating probabilities.

Probability, Part 1 Probability tells us the likelihood of a given event occurring. We normally represent it as a decimal or fraction between 0 and 1, or as a percentage between 0% and 100%.

Probability, Part 1 For example, a weather reporter might say there is a 70% chance, or probability, of rain tomorrow. We could also say P(rain) = 0.7

Probability, Part 1 A probability of 1, or 100%, indicates certainty that a given even will occur. A probability of 0, or 0%, indicates certainty that a given even will not occur.

Probability, Part 1 Example 1: Say you roll a die. What is the probability that you will get a number between 1 and 6? What is the probability you will get a 7? P(1-6) = 1, because we are guaranteed to get one of the numbers from 1 to 6.

Probability, Part 1 Example 1: Say you roll a die. What is the probability that you will get a number between 1 and 6? What is the probability you will get a 7? P(7) = 0, because it is impossible to get a 7 with only one die.

Probability, Part 1 This example was rather simple, but the problems can become much more complex. We need to have a more systematic approach to handle those complex problems. In general, we will want to determine the number of possible outcomes that fit a set of conditions, and compare that to the total possible number of outcomes.

Probability, Part 1 In general, the probability of a given event occurring is equal to the ratio: __Number of Outcomes That Work__ Total Possible Number of Outcomes

Probability, Part 1 Example 2: If you roll a die, what is the probability of getting an even number? ____Number of Even Outcomes____ Total Possible Number of Outcomes

Probability, Part 1 Example 2: If you roll a die, what is the probability of getting an even number? Possible Outcomes: 1, 2, 3, 4, 5, 6 So there are 6 total possible outcomes.

Probability, Part 1 Example 2: If you roll a die, what is the probability of getting an even number? Even Outcomes: 2, 4, 6 There are 3 even outcomes.

Probability, Part 1 Example 2: If you roll a die, what is the probability of getting an even number? P(even) ==

Probability, Part 1 Again, this example was rather simple. Let's look at a couple other examples that require more thought.

Probability, Part 1 Example 3: You are taking a 5-question matching quiz; you are asked to match 5 vocabulary terms with 5 definitions. Alas, you are ill-prepared for the quiz, and end up totally guessing on the whole thing. What is the probability that by guessing you will score 100%?

Probability, Part 1 Example 3: We can view this as a permutation problem to determine the total possible number of ways to answer the quiz. In a sense, you are being asked to arrange 5 definitions in order. The number of ways to do this is P(5, 5) = 5 nPr 5 = 120.

Probability, Part 1 Example 3: The outcomes that “work” in this case are the outcomes that will give you a score of 100%. There is only 1 such outcome. Hence, the probability of scoring 100% is 1 = , or about 0.83%. 120

Probability, Part 1 Example 4: You are taking a 10 question True/False quiz. You are so unsure of the questions that you decided to flip a coin to help you answer each one. (If you get heads, you answer true; tails you answer false.) In other words, you're answering completely at random. What is the probability you will score exactly 70%?

Probability, Part 1 Example 4: 10 question True/False Quiz First, let's figure out the total number of ways to answer the quiz. The first question has two options: true or false. So does the second question, and the third,... So the total possible number of outcomes is 2*2*2*2*2*2*2*2*2*2 = 2 10 = 1,024.

Probability, Part 1 Example 4: 10 question True/False Quiz Now let's figure out the number of outcomes that “work,” or in this case that give you a score of 70%. Assuming there is no partial credit, this means getting 7 out of the 10 questions correct. It doesn't matter which 7 questions you get correct, though—it could be 1-7, 4-10, 1-5 and 8-9, etc.

Probability, Part 1 Example 4: 10 question True/False Quiz In other words, we need the number of combinations of 7 correct answers out of 10 problems. (Why combinations rather than permutations?) The number of ways to score 70% is C(10, 7) = 10 nCr 7 = 120.

Probability, Part 1 Example 4: 10 question True/False Quiz The probability of getting 70% is then __120__ 1,024 = 0.117, or about 11.7%.

Probability, Part 1 In the previous example we used the multiplication principle together with combinations. In some instances we will use the addition principle in concert with combinations or permutations. (Recall that a key word for the addition principle is frequently “or.”)

Probability, Part 1 Example 5: You are again completely guessing on a 10 question True/False quiz. What is the probability that you will score at least 70%?

Probability, Part 1 Example 5: 10 question True/False Quiz Again assume that no partial credit is given. In this case, the outcomes that “work” are scores of 70% or 80% or 90% or 100%. We need to determine these possibilities and then add them together.

Probability, Part 1 Example 5: 10 question True/False Quiz We already calculated the number of ways to get 7 out of 10 questions correct: C(10, 7) = 10 nCr 7 = 120.

Probability, Part 1 Example 5: 10 question True/False Quiz Consider also the number of ways to get 8, 9, or 10 correct: C(10, 8) = 10 nCr 8 = 45. C(10, 9) = 10 nCr 9 = 10. C(10, 10) = 10 nCr 10 = 1.

Probability, Part 1 Example 5: 10 question True/False Quiz Now add together the number of ways to get 7, 8, 9, or 10 questions correct = 176 This gives us a total of 176 outcomes that work.

Probability, Part 1 Example 5: 10 question True/False Quiz The total possible number of ways to answer the quiz remains 1,024, as we already calculated in Example 4.

Probability, Part 1 Example 5: 10 question True/False Quiz This gives us a probability of scoring at least 70% on the quiz as: 176__ 1,024 = 0.172, or about 17.2%.

Probability, Part 1 The Addition Principle for Mutually Exclusive Events When considering 2 events, A and B, we can find the probability of one or the other occurring by simply adding together their individual probabilities, with one adjustment.

Probability, Part 1 The Addition Principle for Mutually Exclusive Events When considering 2 events, A and B, we can find the probability of one or the other occurring by simply adding together their individual probabilities, with one adjustment. We need to subtract the overlap between the 2 events. P(A or B) = P(A) + P(B) – P(A and B)

Probability, Part 1 The Addition Principle for Mutually Exclusive Events In a case where A and B are mutually exclusive, both cannot occur at the same time. As a result, P(A and B) = 0, and our equation simplifies to P(A or B) = P(A) + P(B)

Probability, Part 1 The Addition Principle for Mutually Exclusive Events Confused yet? Let's look at some examples to help illustrate.

Probability, Part 1 The Addition Principle for Mutually Exclusive Events Example 6: A car is being given away at a high school graduation. What is the probability that a football player or band member will win the car, if among the 400 graduates there are 40 football players and 50 band members, including 5 that participated in both football and band?

Probability, Part 1 The Addition Principle for Mutually Exclusive Events Example 6: Car Giveaway Out of 40 football players, 5 were also in band, so there were 35 who played football but were not in band. Out of 50 band members, 5 also played football, so there were 45 who were in band but did not play football.

Probability, Part 1 The Addition Principle for Mutually Exclusive Events Example 6: Car Giveaway Football Band

Probability, Part 1 The Addition Principle for Mutually Exclusive Events Example 6: Car Giveaway Football Band P(Football or Band) = P(Football) + P(Band) – P(Football and Band) = 40_ + 50_ - 5 = 85 =

Probability, Part 1 The Addition Principle for Mutually Exclusive Events Example 7: At the same high school graduation, what is the probability that a football player or cross country runner will win the car, if among the 400 graduates there are 40 football players and 50 cross country runners, and no one participated in both?

Probability, Part 1 The Addition Principle for Mutually Exclusive Events Example 7: Car Giveaway Football 40 CC 50 P(Football or CC) = P(Football) + P(CC) = 40_ + 50_ = 90 =

Probability, Part 1 Conditional Probability:Using the Multiplication Principle We can use the multiplication principle to help determine the probability of two events both occurring (and rather than or): P(A and B) = P(A) * P(B from A) Once again, this may be confusing without an example.

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Example 8: A certain high school has the following student population: What is the probability that a student selected at random will be a male junior? FemaleMaleTotal Fr Soph Jr Sr Total

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Example 8: P(Junior and Male) Consider being a junior as Event A, and being male as Event B. We’re trying to determine P(A and B). We could do this by constructing a tree diagram.

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Example 8: P(Junior and Male) Fr Sr Jr So Label the probabilities for each grade level. These come from the number of students in each grade divided by the total number of students. E.g., P(So) = 300/1000 =

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Example 8: P(Junior and Male) Fr Sr Jr So M M M M F F F F Label the probabilities of being male or female within each grade level.

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Example 8: P(Junior and Male) Fr Sr Jr So M M M M F F F F Follow the path for Juniors and then Males, and multiply the probabilities *.431 =.110

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Example 8: P(Junior and Male) We can get the same result by using the formula: P(A and B) = P(A) * P(B from A)

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Example 8: P(Junior and Male) P(A and B) = P(A) * P(B from A) P(Junior and Male) = P(Junior) * P(Male from Juniors) P(Junior) = 255/1000 =.255 P(Male from Junior) = 110/255 =.431 P(Junior and Male) =.255 *.431 =.110

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Our formula for conditional probabilities becomes a small bit simpler when the events are independent. In this case: P(A and B) = P(A) * P(B)

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Two events are independent if P(B from A) = P(B) or P(A from B) = P(A)

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Consider again our example of the school where Event A is being a Junior, and Event B is being a male. Does P(B from A) = P(B)?

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Consider again our example of the school where Event A is being a Junior, and Event B is being a male. P(B) = P(Male) = 440/1000 =.440 P(B from A) = P(Male from Juniors) = 110/255 =.431 So P(B from A)  P(B)

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Consider again our example of the school where Event A is being a Junior, and Event B is being a male. P(A) = P(Junior) = 255/1000 =.255 P(A from B) = P(Junior from Males) = 110/440 =.250 So P(A from B)  P(A) Being a junior and being male are therefore not independent.

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Example 9: Are being a sophomore and being female independent in the school of Example 8? FemaleMaleTotal Fr Soph Jr Sr Total

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Example 9: Are being a sophomore and being female independent in the school of Example 8? P(B) = P(Female) = 560/1000 =.560 P(B from A) = P(Female from Sophomores) = 168/300 =.560 Because P(B from A) = P(B), being a sophomore and being female are independent in this school.

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Example 9: Are being a sophomore and being female independent in the school of Example 8? In other words, the probability of picking a female from among the sophomores is the same as the probability of picking a female from the school population at large. In contrast, the probability of picking a male from among the juniors is slightly less than the probability of picking a male from the school population at large.

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Example 10: What is the probability that a student chosen at random from the entire school population will be a female sophomore?

Probability, Part 1 Conditional Probability:Using the Multiplication Principle Example 10: What is the probability that a student chosen at random from the entire school population will be a female sophomore? We already established that being female and being a sophomore are independent events in this case, so we can use our simplified formula: P(Female and Sophomore) = P(F) * P(Soph) =.560 *.300 =.168 This makes sense as there are 168 female sophomores out of 1,000 total students in the school.