Solution Activity Set 6

SOLUTION ACTIVITY SET 6

6.1 Situation: In 1998 the American Film Institute created a list of the top 100 American films ever made (http://www.afi.com/tvevents/100years/movies.aspx). Suppose that you and a date gather to watch one of these movies and, to avoid potentially endless debates about a selection, decide to choose a movie at random from the “top 100” list. You will investigate the probability that it has already been seen by at least you or your date. Let A denote the subset of these 100 films that you have seen, so the event A = {films that you have seen}. Similarly define event B for your date. The following 2x2 table classifies each movie according to whether it was seen by you and whether it was seen by your date. The “at random” selection implies that each of the 100 films is equally likely to be chosen; i.e., each has probability 1/100. Thus, the probabilities of these various events can be calculated by counting how many of the 100 films comprise the event of interest.

Date Yes Date No Total You Yes 42 6 48 You No 17 35 52 Total 59 41 100

For example, the table reveals that 42 movies were seen by both you and your date, so P(A ∩ B) = 42/100 = 0.42 (read “the probability of A and B”).

1a. Fill in the marginal totals of the table (the row and column totals). From these totals determine the probability that you have seen a randomly selected film and also the probability that your date has seen the film. (Remember that the film is chosen at random, so all 100 are equally likely.) Answer the following placing the proper event in the ( ). 2 P(You have seen it) = P( A ) = .48

P(Date has seen it) = P( B ) = .59

1b. Translate the following events into set notation using the symbols A and B, complement, union, intersection. Also give the probability of the event as determined from the table above. Fill in the table below with these values. {If you cannot construct the symbol ∩, just use &} Lastly, draw a Venn diagram (on scratch paper, but do not hand in) showing the events of you, your date, and both you and your date seeing the movie.

Event in words Event in set notation Probability Allan and Beth have both seen the film A ∩B P(A ∩B)=.42 Allan has seen the film and Beth has not A ∩Bc P(A ∩Bc)=.06 Beth has seen the film and Allan has not Ac∩B P(Ac ∩B)=.17 Neither Allan nor Beth has seen the film Ac∩Bc P(Ac ∩Bc)=.35 1 2 3c. Determine the probability that you have not seen the film. Do the same for your date. P(You have not seen it) = P( Ac ) = 0.52 P(Your date has not seen it) = P( Bc ) = 0.41 4. 5d. If you had not been given the table, but instead had merely been told that P(A)=.48 and P(B)=.59, would you have been able to calculate P(Ac) and P(Bc)? Explain how. Yes we could compute the probability of the complement of an event using the probability of the event by subtracting the value of the probability of the event from one: P(You have not seen it) = P(Ac) = 1- P(A)=1-.48 = .52 P(Your date has not seen it) =P(Bc) =1- P(B) = 1-.59= .41 e. Calculate the probability that at least either you or your date has seen the film. Using the addition rule: P(A ∪ B) = P(A) + P(B) – P(A ∩ B)=.48+.59-.42=.65 1 2f. Another way to calculate P(A ∪ B) in part e is to use the complement of the event. First, in words what is the complement to the event described in e and then find this probability using the complement. Does this match your answer in e? The complement of the event {You or your date has seen the movie} is the event {neither you nor your date has seen the movie}. In symbols we can describe this event as (A ∪ B)c which equals Ac ∩ Bc. Thus P(A ∪ B)= 1 - P(Ac ∩ Bc) = 1-.35 = .65 1g. If you had not been given the table but instead had merely been told P(A)=.48 and P(B)=.59 and were asked to calculate P(A ∪ B), you might first consider P(A)+P(B). Calculate this sum and compare the result to your answer for e. Are they the same? Is this even a legitimate answer? P(A)+P(B)=1.07 This value does not equal the value of P(A ∪ B) computed above. It is also not a legitimate answer since the probability function always returns a number in the interval [0,1]. 1h. Given the knowledge that you have seen a film, what is the conditional probability that your date has seen it? [Hint: Restrict your consideration to films that you have seen, and ask yourself what fraction of those your date has seen.] 0 This is asking to find P(B|A) = P(A ∩ B)/P(A) = 0.42/0.48 = 0.875 i. How does this conditional probability of your date having seen the film given that you have seen it compare with the (unconditional) probability of your date having seen the film in the first place? Does the knowledge that you have seen the film make it more or less likely (or neither) that your date has seen it? The conditional probability (0.875) is greater than the unconditional probability (0.59) that your date saw the movie in the first place. Knowing that you have seen the movie makes the probability that your date has seen the movie more likely.

6.2 Graduate School Admissions Suppose that you apply to two graduate schools A and B, and that you believe your probability of acceptance by A to be 0.7, your probability of acceptance by B to be 0.6, and your probability of acceptance by both to be 0.5. a. Are the events {acceptance by A} and {acceptance by B} independent? Explain. {Hint: look at the definition for independence from the lecture notes or in the Probability Rules table on page 223.} If two events are independent the probability that both events occur equals the product of their individual probabilities. That is, P(A ∩ B) = P(A)*P(B). In this case, P(A)*P(B) = 0.7*0.6 = 0.30, and P(A ∩ B) = 0.5. Since these are not equal, then A and B are not independent. b. Determine the conditional probability of acceptance by B given acceptance by A? How does it compare to the (unconditional) probability of acceptance by B? P(B|A) = P(A ∩ B)/P(A) = 0.5/0.7 = 0.71 c. What is the probability that you are accepted by at least one of the two schools? P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.7 + 0.6 – 0.5 = 0.8

6.3 Just for fun! Back in the Seventies there was a game show called “Let’s Make a Deal”. The rules were simple: a contestant was selected from the crowd and offered the chance to open one of 3 doors. Behind one door was a prize (e.g. money, new car), while behind the other two doors were gag prizes (e.g. a donkey, barrel of popcorn). After making their choice of door the game show host would open one of the non-selected doors which was always one that did not contain the prize. The contestant was then asked if he or she would like to switch from their first choice to the remaining unopened door or stay with their original choice. If you were the contestant, what choice would you make? That is, given that you now have two doors unopened, one with a prize behind it and the other without, which gives you a better chance of winning: switching doors or staying with your original choice? After you have thought about this, please visit the following website and try it out!

{You may have to click the refresh button - after opening the page} http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

You should switch! The website provides a nice explanation as to why.