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##### Probability II

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**Denoted by P(Event)**Probability This method for calculating probabilities is only appropriate when the outcomes of the sample space are equally likely.**The relative frequency at which a chance experiment occurs**Flip a fair coin 30 times & get 17 heads Experimental Probability**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. Law of Large Numbers**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 Basic Rules of Probability**Rule 3.Complement**• For any event E, • P(E) + P(not E) = 1**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 PWSH? A randomly selected student is female - What is the probability she plays football for PWSH? Independent Independent Not independent**Rule 4.Multiplication**If two events A & B are independent, General rule:**What does this mean?**Independent? Yes**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?**Yes No**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) • (General) If two events E & F are not disjoint, • P(E or F) = P(E) + P(F) – P(E & F)**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. 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? yes P(H or N or T) = .25 + .18+ .14 = .57 P(not (H or N or T) = 1 - .57 = .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) = .4 + .3 -.1 = .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? NO, independent events cannot be 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! P(A or B) = P(A) + P(B) – P(A & B) If independent, multiply How can you find the probability of A & B? P(A or B) = .45 + .35 - .45(.35) = 0.6425**Coin is TAIL**Roll is LESS THAN 3 H5 T6 H1 T2 T4 H4 T3 H2 T1 T5 H6 H3 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? Sample Space H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6 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! P (heads or even) = P(tails) + P(less than 3) – P(tails & less than 3) = 1/2 + 1/3 – 1/6 = 2/3**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 & DC) or P(DC & 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 & DC) or P(DC & 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(DC & DC) .0975**Dr. Pepper**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) 1 - .96 = .4686**A probability that takes into account a given condition**Rule 7: Conditional Probability**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**Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 12) What is the probability that the driver is a student?**Probabilities from two way tables**Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 13) What is the probability that the driver drives a European car?**Probabilities from two way tables**Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 14) What is the probability that the driver drives an American or Asian car? Disjoint?**Probabilities from two way tables**Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 15) What is the probability that the driver is staff or drives an Asian car? Disjoint?**Probabilities from two way tables**Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 16) What is the probability that the driver is staff and drives an Asian car?**Probabilities from two way tables**Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 17) If the driver is a student, what is the probability that they drive an American car? Condition**Probabilities from two way tables**Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 18) 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 = 0.048 P(inexperienced|returned) = 2.4/4.8 = 0.5