2.  Denoted by P(Event) This method for calculating probabilities is only appropriate when the outcomes of the sample space are equally likely.

3.  The relative frequency at which a chance experiment occurs ◦ Flip a fair coin 30 times & get 17 heads

4.  As the number of repetitions of a chance experiment increase, the difference between the relative frequency of occurrence for an event and the true probability approaches zero.

5. Rule 1. Legitimate Values For any event E, 0 < P(E) < 1 Rule 2. Sample space If S is the sample space, P(S) = 1

6. Rule 3. Complement For any event E, P(E) + P(not E) = 1

7.  Two events are independent if knowing that one will occur (or has occurred) does not change the probability that the other occurs ◦ A randomly selected student is female - What is the probability she plays soccer for FHS? ◦ A randomly selected student is female - What is the probability she plays football for FHS? Independent Not independent

8. Rule 4. Multiplication If two events A & B are independent, General rule:

9. Independent? Yes What does this mean?

10. Ex. 1) 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?

11. Independent? YesNo

12. Ex. 2) There are seven girls and eight boys in a math class. The teacher selects two students at random to answer questions on the board. What is the probability that both students are girls? Are these events independent?

13. Ex 6) Suppose I will 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) Read “probability of B given that A occurs” P(Spade & Spade) = 1/4 · 12/51 = 1/17 The probability of getting a spade given that a spade has already been drawn.

14. Rule 5. Addition If two events E & F are disjoint, P(E or F) = P(E) + P(F) (General) If two events E & F are not disjoint, P(E or F) = P(E) + P(F) – P(E & F)

15. What does this mean? Mutually exclusive? Yes

16. Ex 3) 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, P(T) =.14. What is the probability that the next car sold is a Honda, Nissan, or Toyota? Are these disjoint events? P(H or N or T) = P(not (H or N or T) = yes.25 +.18+.14 =.57 1 -.57 =.43

17. Mutually exclusive? Yes No

18. Ex. 4) 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) =.4 +.3 -.1 =.6

19. Mutually exclusive? Yes No Independent? Yes

20. Ex 5) If P(A) = 0.45, P(B) = 0.35, and A & B are independent, find P(A or B). Is A & B disjoint? If A & B are disjoint, are they independent? Disjoint events do not happen at the same time. So, if A occurs, can B occur? 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) =.45 +.35 -.45(.35) = 0.6425 If independent, multiply

21. H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6 If a coin is flipped & a die rolled at the same time, what is the probability that you will get a tail or a number less than 3? T1 T2 T3 T4 T5 T6H1 H3 H2 H5 H4 H6 Coin is TAILRoll is LESS THAN 3 P (heads or even) = P(tails) + P(less than 3) – P(tails & less than 3) = 1/2 + 1/3 – 1/6 = 2/3 Flipping a coin and rolling a die are independent events. Independence also implies the events are NOT disjoint (hence the overlap). Count T1 and T2 only once! Sample Space

22. 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 exactly one bulb is defective? P(exactly one) = P(D & D C ) or P(D C & D) = (.05)(.95) + (.95)(.05) =.095

23. Ex. 8) 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) = P(D & D C ) or P(D C & D) or (D & D) = (.05)(.95) + (.95)(.05) + (.05)(.05) =.0975

24.  A probability that takes into account a given condition

25. Ex 10) 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?

26. Ex 11) 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?

27. StuStaffTotal American107105212 European331245 Asian5547102 Total195164359 12) What is the probability that the driver is a student?

28. StuStaffTotal American107105212 European331245 Asian5547102 Total195164359 13) What is the probability that the driver drives a European car?

29. StuStaffTotal American107105212 European331245 Asian5547102 Total195164359 14) What is the probability that the driver drives an American or Asian car? Disjoint?

30. StuStaffTotal American107105212 European331245 Asian5547102 Total195164359 15) What is the probability that the driver is staff or drives an Asian car? Disjoint?

31. StuStaffTotal American107105212 European331245 Asian5547102 Total195164359 16) What is the probability that the driver is staff and drives an Asian car?

32. StuStaffTotal American107105212 European331245 Asian5547102 Total195164359 17) If the driver is a student, what is the probability that they drive an American car? Condition

33. StuStaffTotal American107105212 European331245 Asian5547102 Total195164359 18) What is the probability that the driver is a student if the driver drives a European car? Condition

34. Example 19: 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 or 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 = 0.048 P(inexperienced|returned) = 2.4/4.8 = 0.5 Retu rned Not retur ned TotalTotal Experie nced 2.477.68080 Inexperi enced 2.417.62020 Total4.895.2100100

35. ReturnedNot returned Total Experienced2.477.680 Inexperienced2.417.620 Total4.895.2100