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Probability

Probability. Sample space. The set of all possible outcomes of a chance experiment Roll a die S={1,2,3,4,5,6} Pick a card S={A-K for ♠, ♥ , ♣ & ♦ }

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Probability

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  1. Probability

  2. Sample space The set of all possible outcomes of a chance experiment Roll a die S={1,2,3,4,5,6} Pick a card S={A-K for ♠, ♥, ♣ & ♦} We want to know the size of the sample space and we also want to know if each of the outcomes is equally likely. In the examples above, assuming the die is fair and the cards are shuffled, all outcomes are equally likely.

  3. Event • Any subset of the sample space • Rolling a prime number with a single die • Call the event A. Then A can be defined A= {2,3,5} • Pick a spade from a deck of 52 cards and call that event B. B={A♠, 2♠, 3♠, 4♠, 5♠, 6♠, 7♠, 8♠, 9♠, 10♠, J♠, Q♠, K♠}

  4. The set of all outcomes that are not contained in the event From before, the event A is rolling a prime, A = {2,3,5} Therefore AC = {1,4,6} Complement

  5. Union • A or B • The union is the set of all outcomes that are in at least one of the two events • A: Rolling a prime • B: Rolling an even number • Let event E be the union of A and B

  6. Intersection • A and B • The intersection of two events is the set of all outcomes that are in both events • A: Rolling a prime • B: Rolling an even number • Let event E be the intersection of A and B:

  7. If two events have no common outcomes they are said to be mutually exclusive or disjoint. are mutually exclusive. A: Pick a black card from a deck B: Pick a diamond. A and B are mutually exclusive. No card is both black and a diamond. Mutually Exclusive Events

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

  9. The relative frequency at which a chance experiment occurs Flip a fair coin 30 times & get 17 heads The experimental probability is 17/30 or about .567. This is not the same thing as theoretical probability. Experimental Probability

  10. The long-run relative frequency of repeated independent events approaches the true probability as the number of trials increases. Law of Large Numbers

  11. Rule 1.Legitimate Values For any event A, 0 < P(A) < 1 P(A) = 0 implies A cannot occur. P(A) = 1 implies A always occurs Rule 2.Something has to happen If S is the sample space, P(S) = 1 Basic Rules of Probability

  12. Rule 3.Complement rule For any event A, P(AC) = 1 – P(A) If there is a probability of making it to class on time of .95, the complements probability is .05. What does this mean?

  13. Rule 4.Addition If A and B are disjoint, then so the rule simplifies to: This is only the case if A and B are mutually exclusive. Do not use this as a general rule.

  14. Two events are independent if knowing that one will occur (or has occurred) does not change the probability that the other occurs. When two events are independent: Many statistics methods require an independence assumption. Assuming independence does not make it true. Think carefully whether the assumption is reasonable before committing to the multiplication rule. Multiplication Rule

  15. Ex 1) 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? yes P(H or N or T) = .25 + .18+ .14 = .57 P(not (H or N or T) = 1 - .57 = .43

  16. 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) = .4 + .3 – .1 = .6

  17. Ex. 3) 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? Independence is reasonable. P(both defective) = P(B1 defective) times P (B2 defective) = .05(.05)=.0025

  18. Ex. 3 cont.) Same as before. Each bulb has a 5% chance of being defective. What is the probability that at least one of the bulbs works? Let’s use the complement rule. Define event A = {at least one works}. What is AC? AC={neither one works}. We already found P(AC)=.0025. Therefore P(A)=1-.0025=.9975.

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