1. Let it roll.i) A fair, six-sided die is tossed. What is the probability the first 5 occurs on the fourth roll?i) In order for the first 5 to be on the fourth roll, the previous three rolls need to be something other than fives.Pr(first 5 on fourth roll) = Pr(not 5 on first, not 5 on second, not 5 on third, 5 on fourth).Since all rolls are independent, we can break this joint probability into a product of the marginal probabilities.Pr(not 5 on first, not 5 on second, not 5 on third, and 5 on fourth)= Pr(not 5 on first)*Pr(not 5 on second)*Pr(not 5 on third)*Pr(5 on fourth) = (5/6)(5/6)(5/6)(1/6).

ii) Suppose two fair, 6-sided dice are tossed. What is the probability that the sum equals 10 given it exceeds 8?

ii) We want Pr(sum =10 | sum > 8) = Pr(sum = 10 and sum > 8) / Pr(sum > 8)Since a sum of 10 is gerater than 8, the event (sum=10 and sum>8) is equivalent to the event (sum=10). Hence, the above probability simplifies to:Pr(sum=10 | sum>8) = Pr(sum=10) / Pr(sum>8)Using the probability calculations described in part a of the next problem, we get Pr(sum=10) = 3/36.The possible sums greater than 8 include 9, 10, 11, and 12. Adding up the probabilities associated with each of these outcomes, we get Pr(sum > 8)= 10/36.Hence, Pr(sum=10 | sum > 8 ) = (3/36)/(10/36) = 3/10. Notice that this is much larger than 3/36. The extra information that the sum exceeds 8 increases the chance that the sum is 10.

2. Runs of coinsFor fun on Saturday night, you and a friend are going to flip a fair coin 10 times (geek!). Let H be the event that a flip lands with heads showing, and let T be the event that a flip lands with tails showing. Because the coin is fair, assume Pr(H) = Pr(T) = 0.5. Neither of you know how to flip the coin to obtain some desired outcome.You flip HTHHTHTTTH.Your friend flips HHHHHHHTTT.Problems:i) Which sequence is more likely to occur?i) Assuming the coin is fair (i.e., Pr (Heads) = Pr(Tails) = 0.5), and assuming that flips are independent, both sequences are equally likely to occur. This is becausePr(HTHHTHTTTH) = Pr(H) * Pr(T) * Pr(H) * Pr(H) * Pr(T) * Pr(H) * Pr(T) * Pr(T) * Pr(T) * Pr(H) = 0.5 * 0.5 * 0.5 * 0.5 * 0.5 *0.5 * 0.5 * 0.5 * 0.5 * 0.5Pr(HHHHHHHTTT) = Pr(H) * Pr(H) * Pr(H) * Pr(H) * Pr(H) * Pr(H) * Pr(H) * Pr(T) * Pr(T) * Pr(T) = 0.5 * 0.5 * 0.5 * 0.5 * 0.5 *0.5 * 0.5 * 0.5 * 0.5 * 0.5

ii) What is the probability that you will get at least one heads in ten flips?ii) Define the eventsA = heads appears at least one time in ten flipsB = heads appears zero times in ten flips.Because events A and B account for all possible outcomes of 10 flips of the coin,Pr (A) = 1-Pr(B) = 1 - Pr (TTTTTTTTTT) = 1 - (0.5 * 0.5 * 0.5 * 0.5 * 0.5 *0.5 * 0.5 * 0.5 * 0.5 * 0.5)

3. Blood and MarriageBlood comes in four types: O, A, B, and AB. The percentages of people in the United States with each blood type are shown below.Blood Type Percentage-----------------O 46A 40B 10AB 4Problems:i) What is the probability that two people getting married both have blood type O? What assumption are you making?ii) What is the probability that two people getting married both have the same blood type? What assumptions are you making?iii) Do you think that these assumptions are reasonable?For more information on blood types, see the web page linked below.Source:http://www.er4yt.org/Education/Science_A2_Subtype.htmlDefine the events:MO = the man has blood type O. (These are capital Os, not zeros.)WO = the woman has blood type O.Assuming that the man's blood type is independent of the woman's blood type, then the probability that both have blood type O is:P( MO and WO) = P(MO) * P(WO) = .46 * .46 = 0.2116.Similarly, define the events:MA = the man has blood type A.WA = the woman has blood type A.MB = the man has blood type B.WB = the woman has blood type B.MAB = the man has blood type AB.WAB = the woman has blood type AB.Then, the ways for both to have the same blood type include: (MO and WO), (MA and WA), (MB and WB), (MAB and WAB)Assume that the man's blood type is independent of the woman's blood type.We then can compute the probabilities of these events like we did for (MO and WO), then add them to get:Pr (both same blood type) = Pr (MO and WO) + Pr (MA and WA) + Pr (MB and WB) + Pr (MAB and WAB) = .46 * .46 + .40 * .40 + .10 * .10 + .04 * .04 = 0.3832.These assumptions are reasonable provided that people do not look for mates that have similar blood types to them. For example, if people look for someone of the same race or ethnicity, and if blood types are in different proportions for different ethnicities (which they are), then the man's blood type will NOT be independent of the woman's blood type. For example, assuming men with blood type AB look for women that also have AB, then knowing that a man has blood type AB means it is more likely that the woman he is with also will have blood type AB.Moral of the story: assumptions of independence need to be checked thoroughly before they are accepted.

4. Winning at BlackjackA standard deck of playing cards has 52 cards. There are four suits (clubs, diamonds, hearts, and spades), each of which has thirteen numbered cards (2, ..., 9, 10, Jack, Queen, King, Ace).In the game of blackjack, each card is worth an amount of points. Each numbered card is worth its number (e.g., a 5 is worth 5 points); the Jack, Queen, and King are each worth 10 points; and the Ace is either worth your choice of either 1 point or 11 points. The object of the game is to have more points in your set of cards than your opponent without going over 21. Any set of cards that sum greater than 21 automatically loses.Here's how the game is played. You and your opponent are each dealt two cards. Usually the first card for each player is dealt face down, and the second card for each player is dealt face up. After the initial cards are dealt, the first player has the option of asking for another card or not taking any cards. The first player can keep asking for more cards until either he or she goes over 21, in which case the player loses, or stops at some number less than or equal to 21. When the first player stops at some number less than or equal to 21, the second player then can take more cards until matching or exceeding the first player's number without going over 21, in which case the second player wins, or until going over 21, in which case the first player wins.We're going to simplify the game a little andassume that all cards are dealt face up, so that all cards are visible. This is a wimpier game than the face-down one, but it makes for easier probability calculations!Problems:In all these questions, assume your opponent is dealt cards and plays first.i) What is the chance that the first card will be a heartanda Jack?ii) What is the chance that the first card will be a heartora Jack?iii) Given that the first card is a heart, what is the chance that it will be a Jack?iv) Given that the first card is a Jack, what is the chance that it will be a heart?v) Your opponent is dealt a King and a 10, and you are dealt a Queen and a 8. Being smart, your opponent does not take any more cards and stays at 20. What is the chance that you will win if you are allowed to take as many cards as you need?There are 52 cards in a deck, so that the total number of outcomes for the first card is 52. Each is equally likely to be picked.(i) There is only one way to get a Jack and a heart: get the Jack of hearts. Hence, Pr (Jack and heart) = 1/52.(ii) There are 16 ways to get a Jack or a hearts: get one of the thirteen hearts (Ace through King of hearts), or get one of the Jack of clubs, Jack of spades, or Jack of diamonds. Hence, Pr(Jack or hearts) = 16/52. Notice that the Jack of hearts is counted in the thirteen hearts cards, so that I didn't count it again when listing the Jacks.(iii) There are thirteen hearts. There is only one Jack among those 13 hearts: the Jack of hearts. If we know the first card is a hearts, then the chance that it is a Jack is the number of ways to get the Jack of hearts out of the total number of hearts. Hence, Pr(Jack | hearts) = 1/13.(iv) There are four Jacks. There is only one heart among these four Jacks: the Jack of hearts. If we know the first card is a Jack, then the chance that it is a heart is the number of ways to get the Jack of hearts out of the total number of Jacks. Hence, Pr(Jack | hearts) = 1/4.(v) To win, you have to get a 21. There are several ways to get 21: draw a three in one card, draw a two and an Ace in two cards, and draw three Aces in three cards.Getting 21 by drawing a three:Since there are four threes in the deck, and 48 cards remaining after the first four cards are dealt, the chance of getting a 3 is Pr(get a 3) = 4/48.Getting 21 by drawing an Ace first and a 2 second:Pr(get Ace first and a 2 second) = Pr (get a 2 second | get an Ace first) * Pr (get an Ace first) = (4 / 47) * (4/48)The 4/48 is determined as follows. Since there are four aces in the deck, and 48 cards remaining after the first four cards are dealt, the chance of getting an Ace first is 4/48.The 4/47 is determined as follows. Since there are four 2s in the deck, and 47 cards remaining after the first five cards are dealt, the chance of getting a 2 on the next card is 4/47.Getting 21 by drawing a 2 first and an Ace second.By similar logic,Pr(get 2 first and Ace second) = Pr (get Ace second | get 2 first) * Pr (get 2 first) = (4 / 47) * (4/48)Getting 21 by drawing three Aces.There are 4*3*2 = 24 ways to get three aces, and there are (48*47*46) ways to pick three cards. Hence, the probability of picking three aces is 24/(48*47*46).Hence, Pr(win) = 4/48 + (4*4)/(47*48) + (4*4)/(47*48) + (4*3*2)/(47*48*46) = .0907.

5. Sex of childrenDo certain families have a tendency to have babies of the same sex? Let's assume that the sexes of babies are independent (e.g., the sex of the first-born baby does not affect the probability the second-born baby is female) , and that the probability that a baby is born female is 0.5014, which is approximately the current percentage. Assume that this probability is the same regardless of family size (which is an assumption that is not necessarily true).Problems:i) (5 points) What is the chance that a family with 6 children has them born in alternating sex order, with the oldest being a male?ii) (5 points) What is the chance that a family has 6 girls in a row?iii) (5 points) Given that the first three children are boys, what is the chance that the next child will be a girl?An article that analyzes the question of sex tendencies is in the winter 2001 issue of the magazineChance.i) Because the babies' sexes are independent, Pr ( M, F, M, F, M, F) = Pr(M) * Pr(F) * Pr(M) * Pr(F) * Pr(M) * Pr(F) = .4986 * .5014 * .4986 * .5014 * .4986 * .5014ii) Because the babies' sexes are independent, Pr ( F, F, F, F, F, F) = Pr(F) * Pr(F) * Pr(F) * Pr(F) * Pr(F) * Pr(F) = .4986 * .4986 * .4986 * .4986 * .4986 * .4986iii) Because the babies' sexes are independent, Pr ( F | M, M, M ) = Pr (F) = .5014

6. For the future lawyers.In a crime at UNC Chapel Hill, it is determined that the perpetrator is a student who was wearing Duke sweatpants and a Carolina sweatshirt. A student is arrested who was wearing both of these items of clothing on the evening of the crime.The defense provides evidence that shows the probability that a randomly selected student in Chapel Hill is wearing Duke sweatpants is 1/10, and the probability that a randomly selected student in Chapel Hill is wearing a Carolina sweatshirt is 1/5. The prosecutor concludes that the probability that a student is wearing both Duke sweatpants and a Carolina sweatshirt is (1/10)(1/5) = 1/50 = 2%, which is large enough to cause reasonable doubt for the jury.Problem:Do you think the prosecuter's claim about the probability is true or false? Inthree sentences or less, justify your conclusions. Assume the 1/10 and 1/5 are accurate probabilities.The prosecutor's probability calculation is correct only if the events wearing Duke sweatpants and Carolina sweatshirt are independent. This is not likely to be true in Chapel Hill!!! People are either Carolina fans or Duke fans. It is likely that Pr(wear Carolina sweatshirt | wear Duke sweatpants) is pretty small, so that the joint probability of wearing both is pretty small. Hence, the defense's argument for reasonable doubt cannot be based on clothing.When two events A and B are not independent, it isnottrue that Pr (A and B) = Pr (A) * Pr(B).

7. A Lear jet, a mansion, and a big, big, big pool.I know that lotteries are poor bets from a probabilistic point of view. But, sometimes when the jackpot gets up there, I play them anyway. The fantasies alone are worth the $1 for a ticket.Let's calculate the probability of winning the California Daily 3 lottery (there is no lottery in NC). In this lottery, you pick three numbers between 0 and 9. You can pick the same number multiple times, e.g. you can pick 7-7-5 or 7-7-7. You choose to play one of three ways: (i) "straight", which means you win if you pick the three numbers in the exact order that they are drawn, (ii) "box", which means you win if you pick the three numbers correctly in any order, (iii) "straight/box", which means you win if you pick three numbers that win either the straight or the box styles. Prizes for each style are determined from the numbers of people who play the game in that style.For example, say you pick 7-7-5. If the drawn numbers are 7-7-5, you'd win in all three styles of play. If the drawn numbers are 7-5-7, you win the box and straight/box style, but do not win the straight style. If the numbers are 2-0-5, you win nothing.Problem:i) Say you pick 1-2-3. What is the probability of winning straight style? What is the probability of winning box style? What is the probability of winning straight/box style?ii) Say you pick 1-2-2. What is the probability of winning straight style? What is the probability of winning box style? What is the probability of winning straight/box style?iii) Say you pick 2-2-2. What is the probability of winning straight style? What is the probability of winning box style? What is the probability of winning straight/box style?Incidently, because each number is equally likely to come up, and you have to share lottery prizes with others who guessed that number, the best picks are numbers that no one else thinks of writing.Extra Information on LotteriesThe calculations below are more involved and won't be on homeworks, quizzes, or exams. But, you might find it interesting to see how to compute the probability of winning one of the lotteries with the really big jackpots.The big kahuna of the California lotteries is the Super Lotto Plus. (You know it's big because it's not just the Lotto, it's the Super Lotto Plus!) Here is the description from the lottery web site: "SuperLottoPlus is your chance to win millions of dollars! The jackpot ranges from $7 million to $50 million or more. The jackpot rolls over and grows whenever there is no winner. All you have to do is pick five numbers from 1 to 47 and one MEGA number from 1 to 27 and match them to the numbers drawn by the Lottery every Wednesday and Saturday."Source: http://www.calottery.com/games/superlottoplus/superlottoplus.aspFrom this description, you'd think winning was easy, right? Well, let's calculate the probability of getting the correct combination. We'll start by finding the probability of picking the 5 numbers from 1 to 47 correctly.The total number of ways to pick five different numbers between 1 and 47 equals 47*46*45*44*43=184,072,680. Note that different permutations of the same set of five numbers (e.g., 1-2-3-4-5 and 1-2-3-5-4 and 1-2-4-3-5, etc.) are all counted in this 184 billion.Unlike in straight Daily 3, in SuperLotto Plus you don't have to guess the exact order. We have to eliminate duplicate permutations to get the number of distinct combinations of the 5 numbers. To determine the number of combinations of 5 distinct numbers, we use the fact that(number of combinations of 5 distinct numbers) * (number of ways to order each combination of five distinct numbers) = (number of ways to pick five distinct numbers).You can verify this equation by comparing the number of ways to win with 1-2-3 in the straight and box Daily 3 calculations.To save writing, let's label the number of combinations of the 5 numbers asC. For any of these combinations, there are 5*4*3*2*1 = 120 possible ways to order them. Thus, the equation translates to: C * 120 = 184,072,680,so that C = 184,072,680 / 120 = 1,533,939.Now, not only do we have to draw the five numbers correctly, but we have to guess the MEGA number. There are 27 possible choices of the MEGA number. Hence, the total number of winning combinations equals 27 * 1,533,939 = 41,416,353.Therefore, your probability of winning is 1 / 41,416,353. It's very, very , very unlikely that you will win the SuperLotto Plus!!A next step is to determine whether or not the chance at winning the jackpot is worth the $1 cost of a ticket. We'll learn how to do this in a few weeks. Actually, such calculations become pretty complicated when the jackpot can be split among several players.By the way, check out thelottery site on the frequency of each number coming up. This is good data for a final project.There are a total of 1000 numbers going from 000 to 999. Since each digit (0 to 9) is equally likely to occur for all three parts of the number (the 100s place, 10s place, and 1s place), each number must be equally likely to occur.Hence, the chance of picking any particular number is 1/1000.A more formulaic way to calculate this is in two steps:Step 1:Pr(first digit is a 1 and second digit is a 2) = Pr(first digit is a 1) * Pr(second digit is a 2 | first digit is a 1) = 1/10 * 1/10 = 1/100.(Note: what we get on the second digit actually does not depend on what we got on the first digit.)Step 2:Pr(third digit is a 3 and second digit is a 2 and first digit is a 1) =Pr(third digit is a 3 | second digit is a 2 and first digit is a 1) * Pr(second digit is a 2 and first digit is a 1) = 1/10 * (1/100) = 1/1000.i) Straight style: win only when 123 drawn. Pr(win) = 1/1000. Box style: win when any of 123, 132, 213, 231, 312, 321 drawn. Pr(win) = 6/1000. Straight/Box style: win when any of 123, 132, 213, 231, 312, 321 drawn. Pr(win) = 6/1000.ii) Straight style: win only when 122 drawn. Pr(win) = 1/1000. Box style: win when any of 122, 221, 212 drawn. Pr(win) = 3/1000. Straight/Box style: win when any of 122, 221, 212 drawn. Pr(win) = 3/1000.iii) Straight style: win only when 222 drawn. Pr(win) = 1/1000. Box style: win when 222 drawn. Pr(win) = 1/1000. Straight/Box style: win when 222 drawn. Pr(win) = 1/1000.These problems demonstrates independence, which we learn about on Thursday. Two variables are independent when the probability that one occurs is not affected by by whether the other event occurs (e.g., the chance that we get a 2 on the second digit is not affected by what we got on the first digit).

8. The Hardy-Weinberg Law of Genetics

(THIS PROBLEM IS HARDER THAN WHAT WOULD BE ON THE EXAM, BUT IT IS A USEFUL APPLICATION OF PROBABILITY FOR THOSE INTERESTED IN GENETICS)Each hereditary trait in an offspring depends on a pair of genes, one contributed by the father and the other by the mother. A gene is either recessive (denoted bya) or dominant (denoted byA). The hereditary trait isAif one gene in the pair is dominant (AA, Aa, aA), and the trait isaif both genes in the pair are recessive (aa). Suppose that the probabilities of the father carrying the pairsAA, Aa(which is treated as the same asaA), andaaarep0, q0,andr0,respectively, wherep0+ q0+r0= 1.The same probabilities hold for the mother. Also suppose that the gene from each parents is randomly inherited, so that each gene of a pair has a 50% chance of being passed on to the offspring.Assume that matings are random and the genetic contributions of the father and mother are independent.We're going to show that the corresponding probabilities for a first-generation offspring are:p1= (p0+ q0/ 2)2; q1= 2(p0+ q0/ 2) (r0+ q0/ 2) ; r1= (r0+ q0/ 2)2Problems:i) Calculate the probabilities of the offspring getting the gene AA for all sixteen of the possible father/mother pairs, which are shown in the table below.Father Mother----------- AA AA AA Aa AA aA AA aa Aa AA Aa Aa Aa aA Aa aa aA AA aA Aa aA aA aA aa aa AA aa Aa aa aA aa aaii) For each of these pair, calculate the joint probability that the offspring inherits AA and the parents are that pair.iii) Sum up all these probabilities to show that the probability that an offspring gets AA is:p1= (p0+ q0/ 2)2.The other probabilities can be determined in a similar fashion. See the answers for details.TheHardy-Weinberg Lawis that the above probabilities remain unchanged for all future generations of offspring, i.e.,pn= p1,qn= q1,rn= r1for alln>1. This can be shown by mathematical induction.Let the event C = offspring gets AA. For C to happen, the child must get an A from both parents. Let's write all the ways that this can happen:(child is AA and father is AA and mother is AA)(child is AA and father is AA and mother is Aa)(child is AA and father is AA and mother is aA)(child is AA and father is Aa and mother is Aa)(child is AA and father is Aa and mother is aA)(child is AA and father is Aa and mother is AA)(child is AA and father is aA and mother is Aa)(child is AA and father is aA and mother is aA)(child is AA and father is aA and mother is AA)The child cannot be AA when either the mother or father is aa.So, Pr(C) = Pr(child is AA and father is AA and mother is AA) + Pr (child is AA and father is AA and mother is Aa) + Pr (child is AA and father is AA and mother is aA) + Pr(child is AA and father is Aa and mother is Aa) + Pr(child is AA and father is Aa and mother is aA) + Pr(child is AA and father is Aa and mother is AA) + Pr(child is AA and father is aA and mother is Aa) + Pr(child is AA and father is aA and mother is aA) + Pr(child is AA and father is aA and mother is AA)Let's consider the last probability, Pr(child is AA and father is aA and mother is AA). Using the rule we discussed in class, Pr (X and Y) = Pr(Y|X) * Pr(X), we can restate this joint probability as:Pr(child is AA and father is aA and mother is AA) = Pr(child is AA | father is aA and mother is AA) * Pr( father is aA and mother is AA)We can write similar statements for each joint probability in the equation for Pr(C). Assuming that genes are inherited independently, then the chance of getting an A from the father is independent of the chance of getting an A from the mother. Hence,Pr(AA | father is aA and mother is AA) = Pr( get an A from father and an A from mother | father is aA and mother is AA)= Pr( get an A from father | father is aA ) * Pr( get an A from mother | mother is AA)= 0.5 * 1since gene A has a 50% chance of being inherited from the father's gene, aA.Finally, for each father/mother pair, we can compute the Pr(C | father/mother pair)Father Mother----------- AA AA Pr(C | AA, AA) = 1*1 = 1 AA Aa Pr(C | AA, Aa) = 1 * 0.5 = 0.5 AA aA Pr(C | AA, aA) = 1 * 0.5 = 0.5 Aa AAPr(C | Aa, AA) = 0.5 * 1 = 0.5 Aa Aa Pr(C | Aa, Aa) = 0.5 * 0.5 = 0.25 Aa aA Pr(C | Aa, aA) = 0.5 * 0.5 = 0.25 aA AAPr(C | aA, AA) = 0.5 * 1 = 0.5 aA Aa Pr(C | aA, Aa) = 0.5 * 0.5 = 0.25 aA aA Pr(C | aA, aA) = 0.5 * 0.5 = 0.25This answers part 1 of the question.For part 2 of the question, we need to multiply each of the conditional probabilities above by the chance of getting the corresponding father/mother pair.Assuming independence of mating, thenPr(father is aA and mother is AA) = Pr(father is aA) * Pr (mother is AA) = q0* p0.Similar computations can be used to derive that:Father Mother----------- AA AA Pr(AA, AA) = p0* p0. AA Aa Pr(AA, Aa) = p0* q0. AA aA Pr(AA, aA) = p0* q0. Aa AAPr(Aa, AA) = q0* p0. Aa Aa Pr(Aa, Aa) = q0* q0. Aa aA Pr(Aa, aA) = q0* q0. aA AAPr(aA, AA) = q0* p0. aA Aa Pr(aA, Aa) = q0* q0. aA aA Pr(aA, aA) = q0* q0.Adding up the products of the part 1 and part 2, we get:Father Mother----------- AA AA Pr(C | AA, AA) * Pr(AA, AA) = 1 * p0* p0. AA Aa Pr (C | AA, Aa) * Pr(AA, Aa) = 0.50 * p0* q0. AA aA Pr (C | AA, aA) * Pr(AA, aA) = 0.50 * p0* q0. Aa AAPr (C | Aa, AA) *Pr(Aa, AA) = 0.50 * q0* p0. Aa Aa Pr (C | Aa, Aa) * Pr(Aa, Aa) = 0.25 * q0* q0. Aa aA Pr(C | Aa, aA) * Pr(Aa, aA) = 0.25 * q0* q0. aA AAPr(C | Aa, aA) *Pr(aA, AA) = 0.50 * q0* p0. aA Aa Pr(C | aA, Aa) * Pr(aA, Aa) = 0.25 * q0* q0. aA aA Pr(C | aA, aA) * Pr(aA, aA) = 0.25 * q0* q0. -------------------- ---------Pr(C) = p0* p0 + 2 *p0* q0 + 0.5 *q0* q0.This is what we set out to prove.