Probability Written as: P(Event)  Only appropriate if all outcomes are equally likely

Experimental Probability Relative frequency (proportion) of a chance event’s occurrence –Flip a fair coin 30 times, get 17 heads  P(heads) =

Law of Large Numbers With more and more repetitions of a chance experiment, the experimental probability of an event approaches its true probability

Rules of Probability Rule 1: For any event E, 0 < P(E) < 1 Rule 2: If S is the sample space, P(S) = 1 Rule 3: For any event E, P(E) + P(E C ) = 1  P(not E) = 1 – P(E)

Rule 4: General Addition Rule For two events E and F, P(E or F) = P(E) + P(F) – P(E & F)  If E and F are disjoint, P(E & F) = 0

Ex. 1: A large auto center sells cars made by many different manufacturers. Three of these are Honda, Nissan, and Toyota. Suppose that P(H) =.25, P(N) =.18, and P(T) =.14. Are these disjoint events? P(H or N or T) = P(not (H or N or T)) = yes =.57 1 –.57 =.43

Ex. 2: Musical styles other than rock and pop are becoming more popular. A survey of college students finds that the probability they like country music is.40. The probability that they liked jazz is.30 and that they liked both is.10. What is the probability that they like country or jazz? P(C or J) = –.1 =.6

Probabilities from two way tables StuStaffTotal American European Asian Total ) What is the probability that the driver is a student?

Probabilities from two way tables StuStaffTotal American European Asian Total ) What is the probability that the driver drives a European car?

Probabilities from two way tables StuStaffTotal American European Asian Total ) What is the probability that the driver drives an American or Asian car? Disjoint?

Probabilities from two way tables StuStaffTotal American European Asian Total ) What is the probability that the driver is staff or drives an Asian car? Disjoint?

Probabilities from two way tables StuStaffTotal American European Asian Total ) What is the probability that the driver is staff and drives an Asian car?

Probabilities from two way tables StuStaffTotal American European Asian Total ) If the driver is a student, what is the probability that they drive an American car? Condition

Probabilities from two way tables StuStaffTotal American European Asian Total ) What is the probability that the driver is a student if the driver drives a European car? Condition

Independence Two events are independent if knowing that one has occurred does not change the probability that the other occurs –Randomly select a GBHS student  events “female” and “plays soccer for GBHS”  events “female” and “plays football for GBHS” Independent Dependent

Rule 5: General Multiplication Rule For two events A and B,  If A and B are independent,

Ex. 3(a): A certain brand of light bulbs are defective five percent of the time. You randomly pick a package of two such bulbs off the shelf of a store. What is the probability that both bulbs are defective? (Can you assume they are independent?)

Ex. 3(b): A certain brand of light bulbs are defective five percent of the time. You randomly pick a package of two such bulbs off the shelf of a store. What is the probability that exactly one bulb is defective? P(exactly one bad) = P((bad & good) or (good & bad)) = (.05)(.95) + (.95)(.05) =.095

Ex 4: If P(A) = 0.45, P(B) = 0.35, and A & B are independent, find P(A or B). Are A & B disjoint? Disjoint events are dependent! NO, independent events cannot be disjoint P(A or B) = P(A) + P(B) – P(A & B) How can you find the probability of A & B? P(A or B) = –.1575 = If independent, multiply P(A & B) =.45(.35) = If A & B are disjoint, are they independent? Disjoint events cannot happen at the same time. If A occurs, what are the chances of B occurring?

Ex 5: Suppose I pick two cards from a standard deck without replacement. What is the probability that I select two spades? Are the cards independent? NO P(A & B) = P(A) · P(B|A) Probability of B given that A occurs P(Spade & Spade) = 13/52 · 12/51 = 1/17 Probability of getting a spade given that a spade has already been drawn

Ex. 6: The table below shows the results of a survey that asked men and women how they feel about their weight. Are the events “female” and “feels that weight is about right” independent? P(female) = 3021/5876 =.5141 FemaleMaleTotal Underweight About Right Overweight Total P(female | about right) = 1175/2644 =.4444 P(F) ≠ P(F|AR)  Not independent

Ex. 7: A certain brand of light bulbs are defective five percent of the time. You randomly pick a package of two such bulbs off the shelf of a store. What is the probability that at least one bulb is defective? P(at least one bad) = P(bad & good) or P(good & bad) or (bad & bad) = (.05)(.95) + (.95)(.05) + (.05)(.05) =.0975

Rule 6: “At Least One” P(at least 1) = 1 – P(none) Ex. 7 revisited: What is the probability that at least one bulb is defective? P(at least one bad) = 1 – P(none bad) = 1 – P(good & good) =.0975

Ex. 8: For a sales promotion the manufacturer places winning codes under the caps of 10% of all Dr. Pepper bottles. You buy a six-pack. What is the probability that you win something? P(at least one winning code) = 1 – P(no winning symbols) =.4686

Rule 7: Conditional Probability Takes a given condition into account

Ex. 9: In a recent study it was found that the probability that a randomly selected student is a girl is.51 and is a girl and plays sports is.10. If the student is female, what is the probability that she plays sports?

Ex. 10: The probability that a randomly selected student plays sports if they are male is.31. What is the probability that the student is male and plays sports if the probability that they are male is.49?

Ex. 11: Management has determined that customers return 12% of the items assembled by inexperienced employees, whereas only 3% of the items assembled by experienced employees are returned. Due to turnover and absenteeism at an assembly plant, inexperienced employees assemble 20% of the items. Construct a tree diagram and a chart for this data. What is the probability that an item is returned? If an item is returned, what is the probability that an inexperienced employee assembled it? P(returned) = 4.8/100 = P(inexperienced|returned) = 2.4/4.8 = 0.5 ReturnedNot returned Total Experienced Inexperienced Total

Ex. 12: An electronics store sells 2 brands of DVD players. 70% of customers who buy a DVD player choose Brand 1, and the rest choose Brand 2. Of those who buy Brand 1, 20% purchase an extended warranty. Of those who buy Brand 2, 40% purchase an extended warranty. What is the probability that a randomly selected customer who buys a DVD player purchases an extended warranty? If a customer purchases an extended warranty, what is the probability that they chose Brand 2?