1 Probability Part 1 – Definitions * Event * Probability * Union * Intersection * Complement Part 2 – Rules Part 1 – Definitions * Event * Probability * Union * Intersection * Complement Part 2 – Rules

2 Probability Definitions - Event * An event is a specific collection of possible outcomes. * For example, the event ‘A’ could be the event that one Head occurs in two tosses of a coin. * An event is a specific collection of possible outcomes. * For example, the event ‘A’ could be the event that one Head occurs in two tosses of a coin. HHHTTHTT A

3 Probability Definitions - Probability The probability of an event is the ratio of the number of outcomes matching the event description to the number of possible outcomes. P(1 head in 2 tosses) = 2/4 =.5 Probabilities are ALWAYS between 0 and 1. The probability of an event is the ratio of the number of outcomes matching the event description to the number of possible outcomes. P(1 head in 2 tosses) = 2/4 =.5 Probabilities are ALWAYS between 0 and 1.

4 Probability Definitions - Union The union of two events A and B is the event that occurs if either A or B or both occur on a single measurement. * For example, suppose: * A = a car has two doors * B = a car is red * Then, if we randomly select a car from a large parking lot, P(A U B) is the probability that the car is either a two-door model or red or both two-door and red. The union of two events A and B is the event that occurs if either A or B or both occur on a single measurement. * For example, suppose: * A = a car has two doors * B = a car is red * Then, if we randomly select a car from a large parking lot, P(A U B) is the probability that the car is either a two-door model or red or both two-door and red.

5 Probability Definitions – Union Define: C = being female D = being nineteen years old What is P(C U D) for a randomly selected person from among the set of people currently in this room? Define: C = being female D = being nineteen years old What is P(C U D) for a randomly selected person from among the set of people currently in this room?

6 Probability Definitions – Union # of women =________ plus # of 19 year olds =________ minus # of 19 year old women =________ equals # either 19 or a woman or both =________ # of women =________ plus # of 19 year olds =________ minus # of 19 year old women =________ equals # either 19 or a woman or both =________

7 Probability Definition – Union Total # of people in this room = _________ Therefore, P(C U D) =__________ = Total # of people in this room = _________ Therefore, P(C U D) =__________ =

8 Probability Definitions – Intersection The intersection of two events A and B is the event that occurs if and only if both A and B occur on a single measurement. * Suppose we randomly select a person from the set of people currently in this room. With A and B defined as before, the probability that a randomly-selected person is a 19 year old woman is: P(A ∩ B) = The intersection of two events A and B is the event that occurs if and only if both A and B occur on a single measurement. * Suppose we randomly select a person from the set of people currently in this room. With A and B defined as before, the probability that a randomly-selected person is a 19 year old woman is: P(A ∩ B) =

9 Probability Definitions – Complement The complement of an event A is the event that A does not occur. HHTT HTTH A’ A

10 Probability Definitions – Complement P(A’) = 1 – P(A) P(A) + P(A’) = 1 The second equation says, “Either A happens or A doesn’t happen. There are no other possibilities.” That may seem obvious, but keep it in mind on exams. P(A’) = 1 – P(A) P(A) + P(A’) = 1 The second equation says, “Either A happens or A doesn’t happen. There are no other possibilities.” That may seem obvious, but keep it in mind on exams.

11 Probability Part 2 – Rules * Additive Rule * Mutually Exclusive Events * Conditional Probability * Multiplicative Rule * Independence * Additive Rule * Mutually Exclusive Events * Conditional Probability * Multiplicative Rule * Independence

12 Probability Rules – Additive Rule P(A U B) = P(A) + P(B) – P(A ∩ B) * Recall 19 year old women example. * When you add # of 19 year olds to # of women, you count the 19 year old women in both groups – so you count them twice. Subtract that number once as a correction. P(A U B) = P(A) + P(B) – P(A ∩ B) * Recall 19 year old women example. * When you add # of 19 year olds to # of women, you count the 19 year old women in both groups – so you count them twice. Subtract that number once as a correction.

13 Probability Rules – Mutually Exclusive Events Two events are mutually exclusive when P(A ∩ B) = 0 and P(A U B) = P(A) + P(B) Be careful! Note that the first equation uses ∩ while the second one uses U. Two events are mutually exclusive when P(A ∩ B) = 0 and P(A U B) = P(A) + P(B) Be careful! Note that the first equation uses ∩ while the second one uses U.

14 Probability Rules – Conditional Probability When you have information that reduces the set of possible outcomes, you work with new probabilities that are conditional on that new information. *Remember that the # of possible outcomes is the denominator of the ratio that gives probability of an event. *The probability changes on the basis of new information because the numerator stays the same but the denominator decreases. When you have information that reduces the set of possible outcomes, you work with new probabilities that are conditional on that new information. *Remember that the # of possible outcomes is the denominator of the ratio that gives probability of an event. *The probability changes on the basis of new information because the numerator stays the same but the denominator decreases.

15 Probability Rules – Conditional Probability Example: Suppose before class I picked a card at random from a standard deck of cards. What is P(E) if: E = Card I picked is a Club? * Note that this is a question about the ordinary (non- conditional) probability. Example: Suppose before class I picked a card at random from a standard deck of cards. What is P(E) if: E = Card I picked is a Club? * Note that this is a question about the ordinary (non- conditional) probability.

16 Probability Rules – Conditional Probability P(E) = 13/52 =.25(Do you see why?) Now, suppose I tell you that the card I picked is black. What is the conditional probability that the card is a club given that it is black? P(E │black) = 13/26 =.5 * Only 26 cards in a normal deck are black. P(E) = 13/52 =.25(Do you see why?) Now, suppose I tell you that the card I picked is black. What is the conditional probability that the card is a club given that it is black? P(E │black) = 13/26 =.5 * Only 26 cards in a normal deck are black.

17 Probability Rules – Conditional Probability We write: P(A │B) = P(A ∩ B) P(B) We write: P(A │B) = P(A ∩ B) P(B)

18 Probability A B A ∩ B P(A │B) = P(A ∩ B) P(B)

19 Probability Rules – Multiplicative Probability P(A ∩ B) = P(A) * P(B│A) P(A ∩ B) = P(B) * P(A│B) P(A ∩ B) = P(A) * P(B│A) P(A ∩ B) = P(B) * P(A│B)

20 Probability Rules – Multiplicative Probability To see why, begin with the conditional probability formula, and multiply both sides by either P(A) or P(B): P(A │B) = P(A ∩ B) P(B) P(B │A) = P(A ∩ B) P(A) To see why, begin with the conditional probability formula, and multiply both sides by either P(A) or P(B): P(A │B) = P(A ∩ B) P(B) P(B │A) = P(A ∩ B) P(A)

21 Probability Rules - Independence Events A and B are independent if the occurrence of one does not alter the probability of the other. P(A│B) = P(A) P(B│A) = P(B) Events A and B are independent if the occurrence of one does not alter the probability of the other. P(A│B) = P(A) P(B│A) = P(B)

22 Probability Rules – Independence We can now re-write the multiplicative rule for the special case of independent events: P(A ∩ B) = P(A) * P(B) * This is because P(B│A) = P(B) for independent events. We can now re-write the multiplicative rule for the special case of independent events: P(A ∩ B) = P(A) * P(B) * This is because P(B│A) = P(B) for independent events.

23 Probability Probability – Examples 60% of Western students are female. 60% of female students have a B average or better. 80% of male students have less than a B average. a. What is the probability that a randomly selected student will have less than a B average? 60% of Western students are female. 60% of female students have a B average or better. 80% of male students have less than a B average. a. What is the probability that a randomly selected student will have less than a B average?

24 Probability A = Being female B = Having a B average or better A A’.6.4 B B’ B x.6 =.36.6 x.4 =.24.4 x.2 =.08.4 x.8 =.32

25 Probability Probability – Examples a. What is the probability that a randomly selected student will have less than a B average? P(B’) = P(B’ ∩ A) + P(B’ ∩ A’) * Either we get (B’ and A) or we get (B’ and A’). * That is, either our randomly selected student who has less than a B average is a female or he is a male. a. What is the probability that a randomly selected student will have less than a B average? P(B’) = P(B’ ∩ A) + P(B’ ∩ A’) * Either we get (B’ and A) or we get (B’ and A’). * That is, either our randomly selected student who has less than a B average is a female or he is a male.

26 Probability Probability – Examples P(B’) = P(B’ ∩ A) + P(B’ ∩ A’) By Multiplicative Rule: P(B’) = P(B’│A)*P(A) + P(B’│A’)*P(A’) = (.4*.6) + (.8*.4) =.56 P(B’) = P(B’ ∩ A) + P(B’ ∩ A’) By Multiplicative Rule: P(B’) = P(B’│A)*P(A) + P(B’│A’)*P(A’) = (.4*.6) + (.8*.4) =.56

27 Probability Reminder – Multiplicative Probability P(A ∩ B) = P(A) * P(B│A) P(A ∩ B) = P(B) * P(A│B) P(A ∩ B) = P(A) * P(B│A) P(A ∩ B) = P(B) * P(A│B)

28 Probability Probability – Examples b. If we randomly select a student at Western and note that this student has a B or better average, what is the probability that the student is male?

29 Probability A = Being female B = Having a B average or better A A’.6.4 B B’ B x.6 =.36.6 x.4 =.24.4 x.2 =.08.4 x.8 =.32

30 Probability Probability – Examples We know that overall, 40% of students (A’) are male. But among the students who get a B or better average, what proportion are male? The probability of getting a B or better average is.44 (from.36 for women and.08 for men). Thus, P(A’│B) = P(A’ ∩ B) =.08 =.1818 P(B).44 We know that overall, 40% of students (A’) are male. But among the students who get a B or better average, what proportion are male? The probability of getting a B or better average is.44 (from.36 for women and.08 for men). Thus, P(A’│B) = P(A’ ∩ B) =.08 =.1818 P(B).44

31 Probability Probability – Examples You’re on a game show. You’re given a choice of 3 doors you can open. Behind one door is a car. Behind each of the other two doors is a goat. You win what is behind the door you open. You pick a door, but don’t get to open it yet. The host opens another door, behind which is a goat. The host then says to you, “Do you want to stick with the door you chose or switch to the other remaining door?” Is it to your advantage to switch? You’re on a game show. You’re given a choice of 3 doors you can open. Behind one door is a car. Behind each of the other two doors is a goat. You win what is behind the door you open. You pick a door, but don’t get to open it yet. The host opens another door, behind which is a goat. The host then says to you, “Do you want to stick with the door you chose or switch to the other remaining door?” Is it to your advantage to switch?

32 Probability Probability – Examples Yes – you should switch to the other door. The door you originally chose has a 1/3 rd chance of winning the car. The other remaining door has a 2/3 rd chance of winning the car.

33 Probability Probability – Examples Suppose you originally pick Door #1. The door the host opens is indicated by the boldface type below. What you win if you switch is underlined. Door #1Door #2Door #3 GoatGoatCar GoatCarGoat CarGoatGoat Suppose you originally pick Door #1. The door the host opens is indicated by the boldface type below. What you win if you switch is underlined. Door #1Door #2Door #3 GoatGoatCar GoatCarGoat CarGoatGoat

34 Probability Probability - Examples When you picked Door #1, there was a 2/3 rd probability that the car was behind one of the other doors. * That is still true after the host opens one of those other doors. * But since the host knows where the car is, he opens a door that has a goat behind it. So now the 2/3 rd probability is all associated with the one remaining door. * The critical point is that the host has information – so we are dealing with the conditional probability of the car being behind Door #3 GIVEN THAT the host opened Door #2 (after you picked Door #1). When you picked Door #1, there was a 2/3 rd probability that the car was behind one of the other doors. * That is still true after the host opens one of those other doors. * But since the host knows where the car is, he opens a door that has a goat behind it. So now the 2/3 rd probability is all associated with the one remaining door. * The critical point is that the host has information – so we are dealing with the conditional probability of the car being behind Door #3 GIVEN THAT the host opened Door #2 (after you picked Door #1).