“Baseball is 90% mental. The other half is physical.” Yogi Berra

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

Experimental Probability The relative frequency at which a chance experiment occurs –Flip a fair coin 30 times & get 17 heads

Law of Large Numbers 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.

Basic Rules of Probability Rule 1. Legitimate Values For any event E, 0 < P(E) < 1 A probability is a number between 0 and 1. The probability of rain must be 110%!

Rule 2. Sample space If S is the sample space, P(S) = 1 “Something Has to Happen Rule” The probability of the set of all possible outcomes must be 1. I’m 100% sure you are going to have a boy… or a girl.

If the probability that you get to class on time is.8, then the probability that you do not get to class on time is.2. Rule 3. Complement For any event E, P(E) + P(not E) = 1 P(E) = 1 – P(not E)

Independent 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 SHS? –A randomly selected student is female - What is the probability she plays football for SHS? Independent Dependent

Rule 4. Multiplication If two events A & B are independent, General rule: If the probability of rolling a 5 on a fair dice is 1/6, what is the probability of rolling a 5 three times in a row? P( three 5’s in a row) = (1/6) x (1/6) x (1/6) = 1/216 or

Independent? Yes What does this mean?

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?

Independent? YesNo

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?

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.

Rule 5. Addition If two events E & F are disjoint, P(E or F) = P(E) + P(F) If the probability that a randomly selected student is a junior (A) is.2 and the probability that the student is a senior (B) is.5, what is the probability that the student is either a junior or a senior? P(A υ B) = P(A) + P(B), if A and B are disjoint. P(A υ B) = =.7

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) Probability of owning a MP3 player:.50 Probability of owning a computer:.90 So the probability of owning a MP3 player or a computer is 1.40? Not disjoint events!

What does this mean? Mutually exclusive? Yes

Ex 3) A large auto center sells cars made by many different manufacturers. Three of these are Honda, Nissan, and Toyota. (Note: these are not simple events since there are many types of each brand.) Suppose that P(H) =.25, P(N) =.18, P(T) =.14. Are these disjoint events? P(H or N or T) = P(not (H or N or T) = yes = =.43

Mutually exclusive? Yes No

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) = =.6

Mutually exclusive? Yes No Independent? Yes

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) = (.35) = If independent, multiply

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 TAILSRoll is LESS THAN 3 P (tails 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

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

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

Rule 6. At least one The probability that at least one outcome happens is 1 minus the probability that no outcomes happen. P(at least 1) = 1 – P(none)

Ex. 8 revisited) 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) = 1 - P(D C & D C ).0975

Ex 9) For a sales promotion the manufacturer places winning symbols 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 symbol) = 1 – P(no winning symbols) =.4686

Suppose that 40% of cars in Fort Smith are manufactured in the United States, 30% in Japan, 10% in Germany, and 20% in other countries. If cars are selected at random, what is the probability that it is not US made? P(not US made) = 1 – P(US made) = =.6

Suppose that 40% of cars in Fort Smith are manufactured in the United States, 30% in Japan, 10% in Germany, and 20% in other countries. If cars are selected at random, what is the probability that it is made in Japan or Germany? P(Japanese or German) = P(Japanese) + P(German) = =.4

Suppose that 40% of cars in Fort Smith are manufactured in the United States, 30% in Japan, 10% in Germany, and 20% in other countries. If cars are selected at random, what is the probability that you see two in a row from Japan? P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) =.3 x.3 =.09

Suppose that 40% of cars in Fort Smith are manufactured in the United States, 30% in Japan, 10% in Germany, and 20% in other countries. If cars are selected at random, what is the probability that none of three cars came from Germany? P(no Germany in three) = P(not G) x P(not G) x P(not G) =.9 x.9 x.9 =.729

Suppose that 40% of cars in Fort Smith are manufactured in the United States, 30% in Japan, 10% in Germany, and 20% in other countries. If cars are selected at random, what is the probability that at least one of three cars is US made? P(at least one US in three) = 1 – P(no US in three) = 1 – (.6)(.6)(.6) =.784

Suppose that 40% of cars in Fort Smith are manufactured in the United States, 30% in Japan, 10% in Germany, and 20% in other countries. If cars are selected at random, what is the probability that the first Japanese car is the fourth one you choose? P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (.7) 3 (.3) =.1029

Watch out for: probabilities that don’t add up to 1 don’t add probabilities of events if they are not disjoint don’t multiply probabilities of events if they are not independent don’t confuse disjoint and independent

Rule 7: Conditional Probability A probability that takes into account a given condition

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?

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?

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

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 = P(inexperienced|returned) = 2.4/4.8 = 0.5 ReturnedNot returned Total Experienced Inexperienced Total

“Slump? I ain’t in no slump. I just ain’t hittin.” Yogi Berra