MAT109 · Erich Prisner · Franklin College · 2007-2009

Exercises and Projects for Chapter 4: Probability

  1. Assume you flip a coin and get $3 if the result is heads, and have to pay $2 if the result is tails. What is the expected value?
    Heads and tails are equally likely with probability 1/2 each. Therefore the expected value equals (1/2)·3 + (1/2)·(-2) = 1/2.
  2. You face 5 envelopes, containing $0, $1000, $2000, $3000, and $4000. You randomly choose one of them. What is the expected value?
    All five outcomes are equally likely, and have therefore a probability of 1/5. Therefore the expected values for the money is (1/5)·0 + (1/5)·1000 + (1/5)·2000 + (1/5)·3000 + (1/5)·4000 = it is (0+1000+2000+3000+4000)/5 = 10000/5 = 2000.
  3. You have five $1-bills, three $5-bills, and one $10 bills in your pocket. You randomly choose one of them to give the taxi driver a tip. What is the expected value?
    Each of the nine bills is equally likely, a probability of 1/9. Since five bills are one-dollar bills, the probability to draw one of them is 5/9. In the same way, the probability of drawing a $5 bill is 3/9, whereas the probability of drawing the $10 bill is 1/9. Therefore the expected values is (5/9)·1 + (3/9)·5 + (1/9)·10 = 30/9 = 3.33 dollars.
  4. You want to buy a used car tomorrow, for this reason you went to your bank and took $8000 in cash. In the evening there is a football game which you want to attend, you already have the ticket worth $15. You read in the newspaper that during the last game, 5 out of 80000 spactators were stolen their money. You are living in your appartment since 10 years and never had a burglary. What do you do? 1) Stay at home and watch your money, or 2) go to the game and leave the money at home, or 3) go to the game and take the money with you? Explain.
    1. If you stay at home and watch the money, the $15 for the ticket are lost.
    2. I would estimate that the probability for a burglary is less than 1/3650, since you had so many days at home without a burglary, then the expected value when leaving the money at home is between (1/3650)·(-8000) and 0, thus between 0 and -2.4.
    3. The probability that the money will be stolen at the game is about 5/80,000. Therefore the expected value equals (5/80,000)·(-8000)=-1/2.
      4. Overall I would choose option 2 or 3, but not option 1.
  5. a) Two cards are selected randomly from a shuffled 52-card deck. Someone tells you that the first card is black. What is the probability that the second card is black too?
    b) Again two cards are selected randomly from a shuffled 52-card deck. Someone tells you that at least one of the two cards is black. What is the probability that both cards are black?
    a) Since there are only 25 black cards left in the 51 card stack after the first card, a black one, has been taken, the probability is 25/51, slightly less than 1/2.
    b) What are the simple events? All combinations of two cards where at least one of them is black. Thus, out of all the (52·51)/2 combinations of two cards, the (26·25)/2 red-red combinations are excluded. Therefore we have (52·51)/2 - (26·25)/2 simple events. How many of these simple events considered form the event "two black cards"? Obviously (26·25)/2 of them. Therefore the probability of two black cards if we know that at least one of the cards is black is ((26·25)/2)/((52·51)/2 - (26·25)/2) = (26·25)/(52·51 - 26·25) = 25/77, about 1/3, much less than in (a).

  6. SENATE RACE(compare [Kockesen]) An incumbent senator decides first whether to run an expensive ad campaign for the next election. After that, the challenger decides whether to enter the race or not. The chances for the senator to win are 5/6 with the ad campaign, and 1/2 without. The value of winning the election is 2, of losing -0.5, and the cost of the add campaign is 1. (all values in Million dollars). Analyse the game.


    According to backward induction done in the graph, the senator chooses the ad campaign, and the challenger decides not to run. The expected payoffs for senator and challenger are 1 respectively 0.

  7. a) You put $100 on "red" in European roulette. If you lose, the bet is gone. If you win, you get your bet back, plus additional $100. How much total win or loss would you expect?
    b) You put $10 on the numbers 1, ... 12. If you lose, you lose the bet. If you win, you get your bet back plus additional $20. How much total win or loss would you expect?
    c) You put $5 on the number 13. If you lose, you lose the bet. If you win, you get your bet back plus additional $175. How much total win or loss would you expect?
    a) The expected value is 18/37·100 + 19/37·(-100)=-100/37.
    b) The expected value equals 12/37·20 + 25/37·(-10)=-10/37.
    c) The expected value equals 1/37·175 + 36/37·(-5)=-5/37.
  8. You want to insure your yacht, worth $90,000. A total loss may occur with probability 0.005 in the next year, a 50% loss with probability 0.01, and a 25% loss with probability 0.05. What premiums would you have to pay if the insurance company wants to expect a profit of $200 per year?
    The expected value for the insurance company equals the premium x minus the expected money they have to pay you for the loss, which is 0.005·90000 + 0.01·45000 + 0.05·22500. Thus the expected value equals x - 0.005·90000 - 0.01·45000 - 0.05·22500 = x - 450 - 450 - 1125 = 200. Therefore x = 2225.
  9. A random number generator produces a sequence of three digits, where each one of the digits 0, 1, 2, and 3 has equal probability of 1/4. Each digit is generated independently of the others.
    Draw the probability tree for this three-step experiment.
    Find the probability that a sequence
    a) consists of all ones, or
    b) consists of all odd digits.

    Since all probabilities for the random moves are 1/4 in the tree, they are not explicitely given. Each of the 64 outcomes has an equal probability of (1/4)·(1/4)·(1/4)=1/64.
    a) Therefore the probability for the outcome "111" equals 1/64. b) The eight outcomes obeying this description are "111", "113", "131", "311", "133", "313", "331", and "333". Therefore the probability of this event equals 8·(1/64) = 1/8.
  10. Three digits are generated in three rounds. In the first round, the leading digit is selected. It is either 4 or 5, with equal probability. In each further round, a digit is selected among the digits larger or equal than the previously selected one, with equal probability. That means, if the first digit selected was a 5, the next one could only be one of the digits 5, 6, 7, 8, 9, which all have probability 1/5. If let's say 7 is selectd as second digit, then the third one must be one of 7, 8, 9, all having equal probability of 1/3, and so on.
    Draw the probability tree for this three-step experiment.
    Find the probability that a sequence
    a) consists of all even digits, or
    b) consists of all odd digits.

    The outcomes with all digits even are 444, 446, 448 (each one of probability (1/2)·(1/6)·(1/6)), 466, 468 (each one of probability (1/2)·(1/6)·(1/4)), and 488 (of probability (1/2)·(1/6)·(1/2)). Therefore the total probability for such an outcome is the sum of all these values, which is 1/8.
    The outcomes with all digits odd are 555, 557, 559 (each one of probability (1/2)·(1/5)·(1/5)), 577, 579 (each one of probability (1/2)·(1/5)·(1/3)), and 599 (of probability (1/2)·(1/5)·1). Therefore the total probability for such an outcome is the sum of all these values, which is 17/75.
  11. There are eight balls in an urn, identical except the color. Three of them are blue, three of them red, and two of them green. Consider the three-step experiment of a) Draw the Probability Tree or Probability Digraph for this three-step experiment.
    b) How likely is it that two of the drawn balls are blue, and one of them is green?
    c) How likely is it that all three balls have different colors.
    d) How likely is it that all three balls have the same color.

    b) Three outcomes in the tree obey the property, each one with probability 12/336. Therefore the prabability for that event is 36/336=3/28.
    c) blue-red-green, blue-green-red, red-blue-green, red-green-blue, green-blue-red, and green-red-blue, each one having probability 18/336. Therefore the probability for that event is 108/336=9/28.
    d) Two of the outcomes obey the property, having probability 6/336 each, therefore adding up to a probability of 12/336=1/28.
  12. a) Draw a probability tree for all the possible head-tail sequences that can occur when you flip a coin four times.
    b) How many sequences contain exactly two heads?
    c) How many sequences contain exactly three heads?
    d) Draw the probability digraph for the case where you are only interested in how many heads a sequence occurs, not when they occur. That is, you would identify the situations HT and TH, you would identify HHT and HTH and THH, and so on.
    e) What is the probability to get exactly two heads in a sequence of four attempts?
  13. There are eight balls in an urn, identical except the color. Four of them are blue, two of them red, and two of them green. Consider the three-step experiment of a) Draw the Probability Tree or Probability Digraph for this three-step experiment.
    b) How likely is it that two of the drawn balls are red, and one of them is green?
    c) How likely is it that all three balls have different colors.
    d) How likely is it that all three balls have the same color.

    b) ...
    c) ...
    d) ...
  14. There are eight balls in an urn, identical except the color. Four of them are blue, three of them red, and two of them green. Consider the three-step experiment of a) Draw the Probability Tree or Probability Digraph for this three-step experiment.
    b) How likely is it that two of the drawn balls are blue, and one of them is red?
    c) How likely is it that all three balls have different colors.
    d) How likely is it that all three balls have the same color.

    b) ...
    c) ...
    d) ...
  15. A coin is flipped at most 5 times. When a "Tail" shows, the experiment is stopped, otherwise, and if the coin hasn't been flipped more than 4 times, the coin is flipped again.
    a) Draw the probability tree of the multistep experiment.
    b) What is the probability that exaqctly one "Tail" occured?
  16. A coin is flipped at most 5 times. When a "Tail" occurs the second time, the experiment is stopped, otherwise, and if the coin hasn't been flipped more than 4 times, the coin is flipped again.
    a) Draw the probability tree of the multistep experiment.
    b) Draw the probability digraph of the multistep experiment.
    c) What is the probability that exaqctly one "Tail" occured?
  17. You draw 3 cards out of a shuffled 32 cards deck. What is the probability to have a straight--- a sequence of three consecutive cards, like 10, J, Q?
    There are (32·31·30)/6 = 4960 possibilities of having three cards out of 32. There are 64 straights with a 7 as lowest card, 64 with 8 as lowest card, and so on, until 64 straights with Q as lowest card. Therefore there are 6·64= 384 straights. Therefore the probability for a straight equals 384/4960 = 7.74194%.
  18. ** You draw 3 cards out of a shuffled 32 cards deck. What is the probability for a triple---all of them having the same rank? What is the probability to have a pair but not a triple?
    Assume the three cards are drawn in three rounds. The first card could be any. For a triple, the second must have the same rank as the first. Since there are only 3 of this rank left in the deck of 31 cards, the probability for this to happen equals 3/31. The third card must also have the same rank, for which we get a probability of 2/30. Therefore the probability for a triple equals (3/31)·(2/30)= 1/155 = 0.6452%.
    For a pair but not a triple, ... (3/31)+(3/30)-(1/155) = 19.0323%.
  19. You draw 3 cards out of a shuffled 32 cards deck. What is the probability for a flush---all three cards having the same suit?
    Assume the three cards are drawn in three rounds. The first card could be any. For a flush, the second must have the same suit as the first. Since there are only 7 of this rank left in the deck of 31 cards, the probability for this to happen equals 7/31. The third card must also have the same suit, for which we get a probability of 6/30. Therefore the probability for a flush equals (7/31)·(6/30)= 7/155 = 4.5161%.
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Projects
  1. TENNIS: Steffie and Arantxa are playing a tennis set. Recall that a set consists of games and games consist of points. A game is won by the player who has first won at least four points in total and at least two points more than the other player. A set is won by the player who has won at least six games, and at least two more than the other player. Assume each single point is won by Steffie with probability 0.52 and by Arantxa with probability 0.48. How likely is it that Steffie wins a game? How likely is it that Steffie wins the set? Does it matter whether we play with or without tie-break?

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  2. Project: FINAL EXAM: This game is played by three players: The teacher, student X and student Y. Of course, there are also 20 other students, but they don't make decisions and therefore do not really participate in the game. Student X makes the first move by waering his cap to class or not. Seeing this, Student Y has the same option, cap or not. Seeing the two students with or without caps, the teacher makes the decision to have a special eye (a) on X, or (b) on Y, or (c) the teacher could also relax. Seeing the teacher react, each one of students X and Y decides for himself whether to cheat or not.
    Cheating, when not noticed by the teacher, brings a better grade, a payoff of 1. If the teacher notices cheating, disciplinary measures are taken, a payoff of -6 for the student. Cheating is noticed by the teacher
    • with probability 1/5 if the student has no cap to hide and the teacher has no special eye on the student,
    • with probability 1/10 if the student wears a cap but the teacher has no special eye on the student,
    • with probability 1/3 if the student wears no cap and the teacher has a special eye on the student,
    • with probability 1/6 if the student wears a cap and the teacher has a special eye on the student.
    The teacher has also payoffs. It is the sum of a value of -1 if he or she has to keep an eye on a student, compared to 0 if he or she can relax, and a value of -2 if cheating has occured but gone unnoticed.
    How will the players play? Modify the model if you deem it necessary.

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