Probability

1. Probability

2. Probability • Let • N = the size of a population • X denote a particular event that may occur in the population • nX= number of occurrences of event X in the population Then if we denote P(X) to mean the probability of X being observed in the population of size N we have: P(X) = nX/N

3. Probability Examples Let X denote the appearance of 6 in the roll of a die N = number of possible outcomes = 6 nX = number of outcomes favorable to a 6 = 1 P(rolling 6) = nX /N = 1/6 Let X denote getting exactly 1 head in three tosses of a coin Possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT N = 8 nX = number of outcomes with just one head = 3 (HTT, THT, TTH) P(exactly 1 head) = nX/N = 3/8

4. Probability Relative Frequency Approximation of Probability Count the number of times that event X, nXactually occurs in N trials. Use the rationX/N to estimate the probability of X occurring on any given trial Example Let X be the event of Reggie Miller making a foul shot N = number of previous foul shots attempted = 6679 nX = number of previous successful attempts = 5915 P(success) = nX/N = 5915/6679 = 0.886

5. Probability Law of Large Numbers – If a procedure is repeated a large number of times, the probability of outcome X occurring the next time that the procedure occurs approaches the ratio of times that outcome X occurred in all previous trials. Given the career MLB statistics for two baseball players Chipper Jones Atl 7337 AB 2277 H Ave = 0.310 Alcides Escobar Mil 4 AB 2 H Ave = 0.500 Will Alcides Escobar have a better chance of getting a hit in his next AB than Chipper Jones?

6. Probability Which of these statements most accurately expresses the probability of a successful outcome on the next trial? P(Reggie makes foul shot) = 0.886 P(Chipper gets hit) = 0.310 P(A. Escobar gets hit) = 0.500 What other factors must one consider that can affect the success on any given trial?

7. ~A is used in place of A bar here Probability Let A be any event, and P(A) the probability that event A occurs. Then 0  P(A)  1 If P(A) = 0 The occurrence of event A is impossible If P(A) = 1 The occurrence of event A is a certainty Let ~A be the Complement of A  Event A does not occur P(A) + P(~A) = 1 If P(A) = p, with 0 < p < 1, then P(~A) = 1 - p P(Chipper makes out) = 1 – P(Chipper gets hit) = 1 – 0.310 = 0.690

8. The box represents the population or Universe = 500 The area in the intersection of A and B represents the number of people who havevisited both cities Probability Let the population (Universe) be 500 people attending a lecture Let the set A be the people in this population who have visited Paris Let the set B be the people in this population who have visited Tokyo Assume that the number in A is 280 and the number in B is 79 We can represent this situation with the following diagram The ovals A and B represent the number of people in these two sets B A Let A  B = 27

9. Probability In the previous example we had The size of the population = 500 |A| = the number of people in set A = 280 |B| = the number of people in set B = 79 |A  B| = the number of people in both sets A and B = 27 Let P(A) = the probability that someone randomly selected from the audience has visited Paris P(A  B) = the probability someone randomly selected from the audience has visited Paris or Tokyo P(B/A) = the probability that someone who has visited Paris has also visited Tokyo P(~(A  B)) = probability that someone selected from the audience has visited neither Paris nor Tokyo

10. Number in both A and B/ number in A Probability The Rule for Adding probabilities – Probability that in a single trial event A occurs or event B occurs or they both occur P(A  B) = P(A) + P(B) – P(A  B) Where P(A) = |A|/Nthe number of elements in set A/ size of pop. P(B) = |B|/N and P(A  B) = the number in both A and B From the previous page we obtain P(A) = |A| / N = 280/500 = 0.56 P (A  B) = P(A) + P(B) – P(A  B) = |A|/N +|B|/N - |A  B|/N = 280/500 + 79/500 – 27/500 = 0.56 + 0.158 – 0.054 = 0.664 P(B/A) = | A  B|/|A| = 27/280 = 0.964 P(~ (A  B) ) = 1 – P (A  B) = 1 – 0.664 = 0.336

11. Number who have visited both Number visiting only Tokyo = 79 - 27 Number not visiting Tokyo = 500 - 79 Number Visiting Tokyo Number visiting Paris but not Tokyo = 280-27 P(A  B) = P(A) + P(B) – P(A  B) = (253+52)/ 500 Number visiting neither city Probability Another way to examine the information contained in the previous example is to construct a table Visited Tokyo Yes No Total Yes No Total 27 253 280 52 Visited Paris 168 220 79 421 500 Number who have visited Paris

12. Probability While the addition rule takes into consideration that two (or more) different events could occur in a single trial – Like a single person that has visited either Paris or Tokyo or both being selected from the audience in a game show – The multiplication rule applies to events A and B occurring on successive trials. If the outcome of the second trial is independent of the outcome of the first, then P(A and B) = P(A) * P(B) The probability of tossing head on a second toss of a coin is independent of what face appeared on the first toss P(H and H) = P(H) * P(H) = (1/2) * (1/2) = 1/4

13. 1 1 2 2 3 3 4 4 1 5 5 2 6 6 3 4 5 6 Probability What is the probability of rolling a 7 with a pair of dice? The outcome of the roll of the second die is independent of the outcome of the first. We can illustrate the possible outcomes with a tree diagram Die 2 There are 36 equally likely outcomes (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) Die 1 P(7) = 6 / 36

14. Probability In a population of 500 school children half were randomly selected to receive a flu shot. The other half did not receive a flu shot. Some of the children subsequently came down with the flu. The table below summarizes the results. Got the flu Yes No Total Yes No Total 12 238 250 vaccinated 43 207 250 55 445 500 P(Vaccinated and Gets Flu) = P(F/V)P(V) = (12/250) * (1/2) = 12 / 500

15. Probability If at the beginning of the year the odds makers set the probability of the Red Sox winning the World Series to be 1/15 and the Patriots winning the Super Bowl to be 3/25, what is the probability that both teams will win their championships? Are these two events independent?