7  Practice problems

Author

Karl Gregory


Problem 7.1 (Two dice) Consider rolling two dice and let \[\begin{align} A &= \text{both rolls are at least $3$}\\ B &= \text{both rolls are $3$ or less} \\ C &= \text{the sum of the rolls is $10$ or more}\\ D &= \text{the absolute value of the difference between the rolls is at most $1$}. \end{align}\] Give the following probabilities:

  1. \(P(B)\) 1
  2. \(P(C)\) 2
  3. \(P(D)\) 3
  4. \(P(A \cup B)\) 4
  5. \(P(A \cap B)\) 5
  6. \(P(A \cap B^c)\) 6
  7. \(P( (A \cap B)^c )\) 7
  8. \(P(A^c \cup B^c)\) 8
  9. \(P((A \cup B)^c)\) 9
  10. \(P(A^c \cap B^c)\) 10
  11. \(P(C \cap D)\) 11
  12. \(P(C \cup D^c)\) 12

Problem 7.2 (Marbles) Consider a bag of marbles, \(19\) of which are green, \(25\) of which are blue, and \(6\) of which are red. Moreover, suppose \(9\) of the green marbles are opaque, \(5\) of the blue marbles are opaque, and \(3\) of the red marbles are opaque, and the rest of the marbles are transparent.

  1. Suppose you draw one marble from the bag. Give the probability that you draw:
    1. A red marble.13
    2. A transparent green marble. 14
    3. An opaque marble. 15
    4. A marble that is either blue or opaque or both. 16
  2. Suppose you remove all the opaque marbles from the bag and then draw one marble. Give the probability that you draw:
    1. A green marble. 17
    2. A red or blue marble. 18

Problem 7.3 (Athletes) Suppose you draw \(1\) athlete at random from a group of \(100\) athletes such that: \(30\) swim; \(44\) run; \(9\) swim and run; \(5\) swim, bike, and run; \(11\) swim and bike; \(10\) bike and run but do not swim; and \(35\) only bike. Let \(S\), \(B\), and \(R\) denote the events that the athlete you draw swims, bikes, and runs, respectively. Give the following probabilities:

  1. \(P(S \cup R)\) 19
  2. \(P(S \cap R^c)\) 20
  3. \(P(B)\) 21
  4. \(P(S \cup B)\) 22
  5. \(P( (S \cap R) \cap B^c)\) 23
  6. \(P(S^c \cup R^c)\) 24
  7. \(P((R \cap B) \cup (R \cap B^c))\) 25

Problem 7.4 (Candyland) An iteration of the board game Candyland has the following cards: \(6\) each of red, purple, yellow, blue, orange, and green cards, and then \(4\) each of double red, double purple, double yellow, double blue, double orange, and double green cards. In addition, there is one lollipop, one peppermint, one peanut, and one ice cream card.

  1. If you draw \(5\) cards, one at a time, placing each card back into the deck and shuffling before the next draw, how many sequences of draws are possible? 26

  2. If you draw \(5\) cards, one at a time, without placing each card back before drawing the next, what is the probability that one of your \(5\) cards will be the ice cream cone? 27

Problem 7.5 (Tea party) Three guests to a tea party sit down in three chairs. The host then re-arranges them (if necessary) according to a pre-made seating chart.

  1. What is the probability that all three guests may stay in their chairs? 28

  2. What is the probability that at least one guest may stay in her chair? Hint: Just write down all the sample points. Don’t try to use counting rules.29

Note: If we suppose there are \(N\) guests and \(N\) chairs, we must find the answer using what is called the inclusion-exclusion formula, which YSP has not covered in these notes.

Problem 7.6 (Train) A train to the frontier will have \(4\) passenger cars, \(3\) cattle cars, and \(2\) luggage cars.

  1. Bandits will enter two cars, selected at random. Give the probability that the bandits enter:

    1. Two passenger cars.30
    2. A luggage car and a passenger car.31
    3. At least one cattle car.32

  2. Suppose the train cars are put in random order. With what probability are all the passenger cars connected (with no non-passenger car between any two passenger cars)?33

  3. There are \(14\) head of cattle to be transported in the three cattle cars.

    1. In how many ways can \(5\), \(5\), and \(4\) head of cattle, respectively, be put into the three cattle cars?34
    2. Of the \(14\) head of cattle, \(3\) are Jersey cows. If \(5\), \(5\), and \(4\) of the \(14\) head of cattle are put into the three cattle cars at random, with what probability will the Jersey cows all be placed in the same car?35

  4. Mr.and Mrs. Wilkins and their two daughters and three sons are boarding one of the passenger cars on the next stop.

    1. In how many different orders can the members of the Wilkins family enter the car? 36
    2. In how many of these orders does Mrs. Wilkins precede Mr. Wilkins? 37
    3. In how many of these orders does Mrs. Wilkins and her two daughters precede Mr. Wilkins his three sons? 38

  5. At each of the \(10\) stops along Lil’ Jonnie’s journey on the train, he will decide whether or not to scrawl his initials somewhere at the train station. At how many unique sets of stations can Lil’ Jonnie scrawl his initials?39

Problem 7.7 (Bowling balls) Suppose there are \(5\) bowling balls which are identical except that one is magical and delivers, no matter what, a strike with probability \(3/4\). Suppose you get a strike \(1\) out of \(4\) times on average when using a non-magical bowling ball. You select one of the \(5\) balls at random and send it down the lane…

  1. Give the probability that you get a strike. 40
  2. Given that you got a strike, what is the probability you chose the magic bowling ball? 41
  3. Suppose you choose a ball and with the same ball you get two strikes in a row. What is the probability that you chose the magic ball? 42

Problem 7.8 (Marbles again) Consider a bag of marbles, \(19\) of which are green, \(25\) of which are blue, and \(6\) of which are red. Moreover, suppose \(9\) of the green marbles are opaque, \(5\) of the blue marbles are opaque, and \(3\) of the red marbles are opaque, and the rest of the marbles are transparent. Suppose you draw one marble from the bag and let \(G\), \(B\), and \(R\) be the events that the marble is green, blue, and red, respectively, and let \(O\) be the event that it is opaque.

  1. Find \(P(R|O)\). 43
  2. Find \(P(R|O^c)\). 44
  3. Find \(P(R)\). 45
  4. Check whether \(R\) and \(O\) are independent. 46
  5. Find \(P(B^c|O)\). 47
  6. Find \(P(O|G)\). 48
  7. Find \(P(B \cup G|O)\). 49

Problem 7.9 (Goggles) Suppose you order a new pair of swimming goggles and that they will be manufactured with a defect making them leaky with probability \(1/40\).

  1. What is the probability of receiving a defective pair, and then again receiving a defective pair when you re-order the goggles? 50
  2. What assumption did you make in order to compute your answer to part (a)? 51
  3. What is the probability that you receive a defective pair and then a functioning pair when you re-order the goggles? 52
  4. Give an expression for the probability that you receive \(K\) defective pairs of goggles, where \(K \geq 1\). 53

  1. \(9/36\).↩︎

  2. \(6/36\).↩︎

  3. \(16/36\).↩︎

  4. \(24/36 = 2/3\).↩︎

  5. \(1/36\).↩︎

  6. \(15/36\).↩︎

  7. \(35/36\).↩︎

  8. \(35/36\), by De Morgan’s Laws.↩︎

  9. \(12/36\).↩︎

  10. \(12/36\), by De Morgan’s Laws.↩︎

  11. \(4/36\).↩︎

  12. \(24/36 = 2/3\).↩︎

  13. \(6/50\)↩︎

  14. \(10/50\)↩︎

  15. \(17/50\)↩︎

  16. \(37/50\)↩︎

  17. \(10/33\)↩︎

  18. \(23/33\)↩︎

  19. \(P(S \cup R) = 30/100 + 44/100 - 9/100 = 65/100\).↩︎

  20. \(P(S \cap R^c) = 30/100 - 9/100 = 21/100\).↩︎

  21. \(P(B) = 35/100 + 10/100 + 5/100 + (11-5)/100 = 56/100\).↩︎

  22. \(P(S \cup B) = 30/100 + 56/100 - 11/100 = 75/100\).↩︎

  23. \(P( (S \cap R) \cap B^c) = 4/100 = 1/25\).↩︎

  24. \(P( S^c \cup R^c) = P( (S \cap R)^c) = 1 - P( S \cap R) = 1 - 9/100 = 91/100\).↩︎

  25. \(P((R \cap B) \cup (R \cap B^c)) = P(R) = 44/100\).↩︎

  26. There are \(16\) different cards, so there are \(16^{5} = 1048576\) different sequences of \(5\) draws, with replacement.↩︎

  27. We first count the number of \(5\) card “hands” with the ice cream cone card. The first task is to draw the ice cream cone. There is \(1\) way to do this. The second task is to draw the remaining \(4\) cards from among the remaining \(63\) cards. This can be done in \({ 63 \choose 4 }\) ways. The total number of \(5\) card “hands” is \({64 \choose 5}\). So we have \[ P(\text{get ice cream cone}) = \frac{1 \cdot {63 \choose 4}}{ {64 \choose 5}} = \frac{5}{64}. \]↩︎

  28. There are \(3! = 6\) possible seating arrangements. The guests will choose the correct one themselves with probability \(1/6\).↩︎

  29. If we symbolize the host’s seating arrangement as \(123\) and the possible arrangements the guests may choose as \[ S = \left\{\begin{array}{ccc} 123& 213 & 312 \\ 132& 231 & 321 \end{array}\right\}, \] we see that in \(4\) out of the \(6\) possible arrangements, at least one guest is seated in the right place, so the probability is \(2/3\).↩︎

  30. There are \({9 \choose 2}\) ways to select the two cars. There are \({4 \choose 2}\) ways to choose two passenger cars. So the answer is \[ P(\text{two passenger cars}) = \frac{{4 \choose 2}}{ {9 \choose 2}} = \frac{1}{6}. \]↩︎

  31. We have \[ P(\text{a luggage car and a passenger car}) = \frac{{ 4 \choose 1 } {2\choose 1 }}{ { 9 \choose 2}} = \frac{2}{9}. \]↩︎

  32. We have \[ \begin{align} P(\text{at least one cattle car}) & = P(\text{one cattle car}) + P(\text{two cattle cars}) \\ & = \frac{{3 \choose 1} {6\choose 1}}{{9 \choose 2}} + \frac{{3 \choose 2}}{{9 \choose 2}} \\ & = \frac{1}{12} + \frac{1}{2}\\ & = \frac{7}{12}. \end{align} \]↩︎

  33. First, line up all the passenger cars together; there are \(4!\) ways of doing this. Now, treating the four passenger cars together as a single car (since we are not going to separate them), we have \(6\) “cars” to arrange. This can be done in \(6!\) ways. So we have \(4!\cdot 6!\) ways of arranging the \(9\) cars such that the four passenger cars are together. There are \(9!\) ways to arrange all \(9\) individual cars, so we have \[ P(\text{all passenger cars together}) = \frac{6!4!}{9!} = \frac{1}{21}. \]↩︎

  34. This is a partitioning of the \(14\) head of cattle into groups of \(5\), \(5\), and \(4\). The number of ways to do this is \[ \frac{14!}{5!5!4!} = 252252. \] ↩︎

  35. The number of ways in which the \(3\) Jersey cows can be put together is given by \[ \frac{11!}{2!5!4!} + \frac{11!}{5!2!4!} +\frac{11!}{5!5!1!} = 6930 + 6930 + 2772 = 16632. \] which we get by considering placing them in the first, the second, or the third cattle car. Since there are \(252252\) ways to partition the \(14\) head of cattle into groups of sizes \(5\), \(5\), and \(4\), the probability that the Jersey cows are placed together is \[ \frac{16632}{252252} = 0.06593407. \]↩︎

  36. The family members can enter the car in \(7! = 5040\) orders. ↩︎

  37. Mrs. Wilkins precedes Mr. Wilkins in one half of the orders, that is in \(2520\) orders. ↩︎

  38. Mrs. Wilkins and her two daughters precede Mr. Wilkins and his three sons in \(3!4! = 144\) of the orders. ↩︎

  39. Lil’ Johnnie performs a sequence of \(10\) tasks, each of which can be done in \(2\) ways, so the job can be done in \(2^{10} = 1024\) ways.↩︎

  40. Let \(S\) be the event of a strike and let \(M\) be the event of choosing the magic ball. The problem gives \(P(S|M) = 3/4\), \(P(S|M^c) = 1/4\) and \(P(M) = 1/5\). We have \[ P(S) = P(S|M)P(M) + P(S|M^c)P(M^c) = \frac{3}{4}\cdot \frac{1}{5} + \frac{1}{4} \cdot \frac{4}{5} = \frac{7}{20} \] ↩︎

  41. We have \[ P(M|S) = \frac{P(S|M)P(M)}{P(S|M)P(M) + P(S|M^c)P(M^c)} = \frac{3/20}{7/20} = \frac{3}{7}. \] ↩︎

  42. Let \(S_1\) and \(S_2\) be the event of getting a strike the first and the second time you send the ball down the lane, respectively. We have \(P(S_1\cap S_2| M) = P(S_1|M)P(S_2|M) = 9/16\), assuming independence of \(S_1\) and \(S_2\) given \(M\). Also, \(P(S_1\cap S_2| M^c) = P(S_1|M^c)P(S_2|M^c) = 1/16\). So we have \[ \begin{align} P(M|S_1\cap S_2) &= \frac{P(S_1\cap S_2| M)P(M)}{P(S_1\cap S_2| M)P(M) + P(S_1\cap S_2| M)^cP(M^c)} \\ &= \frac{9/16}{9/16 + 1/16} \\ &= \frac{9}{10}. \end{align} \] ↩︎

  43. It helps to make a table: \[ \begin{array}{l|ccc|c} & \text{Green} & \text{Blue} & \text{Red} & \text{Total} \\\hline \text{Opaque} & 9 & 5& 3& 17 \\ \text{Transparent} & 10 &20 &3 & 33\\ \hline \text{Total} & 19 & 25 & 6 & 50 \end{array} \] Using the table we obtain \(P(R|O) = 3/17\). ↩︎

  44. \(P(R|O^c) = 3/33\). ↩︎

  45. \(P(R) = 6/50\). ↩︎

  46. We have \(P(R) = 6/50 \neq P(R|O) = 3/17\), so \(R\) and \(O\) are not independent. ↩︎

  47. \(P(B^c|O) = 12/17\). ↩︎

  48. \(P(O|G) = 9/19\). ↩︎

  49. \(P(B \cup G|O) = 14/17\). ↩︎

  50. Let \(D_1\) and \(D_2\) be the event that the goggles are defective the first and second time ordered, respectively. Then we have \[ P(D_1 \cap D_2) = P(D_1) P(D_2) = (1/40)^2 = 0.000625. \] ↩︎

  51. The assumption was that \(D_1\) and \(D_2\) were independent events. ↩︎

  52. We have \[ P(D_1 \cap D_2^c) = P(D_1) P(D_2^c) = (1/40)(39/40) = 39 / 1600= 0.024375. \] ↩︎

  53. Letting \(D_i\) be the event that the goggles are defective the \(i\)th time ordered, for \(i=1,\dots,K\), we have \[ P(D_1 \cap \dots \cap D_K) = P(D_1) \times \dots \times P(D_K) = (1/40)^K, \] assuming that \(D_1,\dots,D_K\) are mutually independent. ↩︎