16  Practice problems

Author

Karl Gregory


Problem 16.1 (Sum of two dice rolls) Let \(X\) be the sum of two dice rolls.

  1. Fill out the table below by finding the probabilities \(P(X=x)\) and \(P(X\leq x)\) for \(x=2,\dots,12\). \[ \begin{array}{c|cccccccccccc} x &2&3&4&5&6&7&8&9&10&11&12 \\ \hline P(X = x)& &&&&&&&&&& \\ P(X \leq x)&&&&&&&&&&& \end{array} \] 1

  2. Find \(P(X\geq 10)\) and interpret this value in words. 2

  3. In the board game Settlers of Catan, players roll two dice at the beginning of each turn. The player moves “the robber” if the sum of the rolls is \(7\). On what proportion of turns do you expect this to occur? 3

  4. Find \(\mathbb{E}X\). 4

  5. Find \(\operatorname{Var}X\). 5

  6. Suppose you rolled two dice \(1000\) times, recording each time the sum of the two rolls; afterwards you compute the average of these \(1000\) numbers. To what value would you expect this average to be close? 6

  7. Suppose you rolled two dice \(1000\) times. Each time you take the sum of the rolls, subtract \(7\), square the result, and record this number. Afterwards you compute the average of these \(1000\) numbers. To what value would you expect this average to be close? 7

Problem 16.2 (Four flips of a coin) Suppose a coin is flipped four times and let \(Y\) be the number of heads.

  1. Give the sample space \(S\) for the experiment of flipping a coin four times. 8
  2. Write down the probability distribution of \(Y\) in tabular form \[ \begin{array}{c|cccc} y & \hspace{2in}\\ \hline P(Y = y)& \end{array} \] 9
  3. What is the expected value of \(Y\)? Show calculations. 10
  4. What is the variance of \(Y\)? Show calculations. 11
  5. Regarding \(Y\) as a random variable with a binomial distribution, what are \(n\) and \(p\) for the binomial distribution? 12
  6. Compute \(np\). 13
  7. Compute \(np(1-p)\). 14

Problem 16.3 (Defects) In a shipment of \(100\) parts, \(4\) are defective. The purchaser has time to test only \(10\) randomly selected parts for defects. Let \(X\) be the number of defective parts the purchaser finds.

  1. Give the support \(\mathcal{X}\) of \(X\). 15
  2. What is the probability that the purchaser finds no defective parts? 16
  3. What is the probability that the purchaser finds at least \(1\) defective part? 17
  4. Suppose the shipment has \(d\) defective parts. What is the largest value of \(d\) such that the probability is at least \(0.90\) that the purchaser finds no defects among \(10\) randomly sampled parts? 18

Problem 16.4 (Hearts in a five card hand)  

  1. Draw \(5\) cards from a \(52\)-card deck and let \(X\) be the number of hearts in your hand. Tabulate the distribution of \(X\) in the format \[ \begin{array}{c|c} x & \hspace{2in}\\ \hline P(X = x)& \end{array} \] 19

  2. Draw a single card from a \(52\)-card deck and check if it is a heart. Then shuffle it back into the deck. Do this five times and let \(Y\) be the number of hearts you drew. Tabulate the distribution of \(Y\) in the format \[ \begin{array}{c|c} y & \hspace{2in}\\ \hline P(Y = y)& \end{array} \] 20

  3. Shuffle ten 52-card decks together into a single deck. Draw a five-card hand and let \(V\) be the number of hearts in your hand. Tabulate the distribution of \(V\) in the format
    \[ \begin{array}{c|c} v & \hspace{2in}\\ \hline P(V = v)& \end{array} \] 21

  4. Under what circumstance is the difference between sampling with replacement and sampling without replacement negligible? 22

  1. The probability distribution of \(X\) is given by \[ \begin{array}{c|cccccccccccc} x &2&3&4&5&6&7&8&9&10&11&12 \\ \hline P(X = x)&1/36 & 2/36 & 3/36 & 4/36 & 5/36 & 6/36&5/36& 4/36& 3/36& 2/36 &1/36 \\ P(X \leq x)&1/36 & 3/36 & 6/36 & 10/36 & 15/36 & 21/36 & 26/36& 30/36& 33/36& 35/36 &36/36 \end{array} \]↩︎

  2. It is \(6/36 = 0.167\), so in every \(36\) rolls, we expect to see \(6\) for which the sum is greater than or equal to \(10\). ↩︎

  3. If many many turns were taken, the expected proportion of turns on which the robber is moved is equal to \(6/36=1/6=0.167\).↩︎

  4. We obtain \[ \mathbb{E}X = \frac{2\cdot 1 + 3\cdot2 + \cdots + 12\cdot1}{36} = \frac{252}{36} = 7. \] ↩︎

  5. \[ \operatorname{Var}X = \frac{(2 - 7)^2\cdot 1 + (3-7)^2\cdot2 + \cdots + (12-7)^2\cdot1}{36} = \frac{210}{36} = 5.833333. \] ↩︎

  6. We would expect this number to be close to the expected value of \(X\), which is \(7\). ↩︎

  7. We would expect this number to be close to the variance of \(X\), which is \(5.833333\). ↩︎

  8. The sample space is \[ S = \left\{\begin{array}{ccccc} HHHH&HHHT&HHTT & &\\ &HHTH&HTTH & &\\ &HTHH&TTHH &TTTH & \\ &THHH&THHT &TTHT & \\ & &THTH &THTT & \\ & &HTHT &HTTT & TTTT \end{array}\right\} \] ↩︎

  9. We have \[ \begin{array}{c|ccccc} y & 0 & 1 & 2 & 3 & 4\\ \hline P(Y = y)& 1/16 & 4/16 & 6/16 & 4/16 & 1/16\\ P(Y \leq y)&1/16 & 5/16 & 11/16 & 15/16 & 16/16 \end{array} \] ↩︎

  10. \(\mathbb{E}Y = 2\). ↩︎

  11. \(\operatorname{Var}Y = 1\) ↩︎

  12. They are \(n=4\) and \(p=1/2\). ↩︎

  13. We have \(np = 4(1/2) = 2\). ↩︎

  14. We have \(4(1/2)(1-1/2) = 1\). ↩︎

  15. The support is \(\mathcal{X} = \{0,1,2,3,4\}\). ↩︎

  16. We have \[ P(X = 0) = \frac{{ 4\choose 0 }{ 96 \choose 10}}{ { 100 \choose 10}} = 0.6516305. \] ↩︎

  17. We have \[ P(X \leq 1) = 1 - P(X = 0) = 1 - 0.6516305 = 0.3483695. \] ↩︎

  18. We find that with \(d = 1\) defective part, the probability that the purchaser finds no defects in \(10\) randomly sampled parts is \({1\choose 0}{ 99 \choose 10}/{ 100 \choose 10} = 0.90\). With \(d = 2\) defective parts, the probability is \({2\choose 0}{ 98 \choose 10}/{ 100 \choose 10} = 0.8090909\). Since the probability continues to get smaller as \(d\) increases, the answer is \(d=1\). ↩︎

  19. We may use the R code choose(13,0:5)*choose(52 - 13,5 - 0:5)/choose(52,5) to obtain \[ \begin{array}{c|cccccc} x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline P(X = x)& 0.2215 & 0.4114 & 0.2743 & 0.0815 & 0.0107 &0.0005 \end{array} \] ↩︎

  20. We may use the R code dbinom(0:5,5,1/4) to obtain \[ \begin{array}{c|cccccc} y & 0&1&2&3&4&5 \\ \hline P(Y = y)& 0.2373 &0.3955& 0.2637 &0.0879 &0.0146 &0.0010 \end{array} \] ↩︎

  21. We may use the R code choose(130,0:5)*choose(520 - 130,5 - 0:5)/choose(520,5) to obtain \[ \begin{array}{c|cccccc} v & 0 & 1 & 2 & 3 & 4& 5 \\ \hline P(V = v)& 0.2358 & 0.3970 & 0.2647 & 0.0873 & 0.0143 & 0.0009 \end{array} \] ↩︎

  22. When the population is large, sampling with replacement and sampling without replacement will give very similar results, since the composition of a large population changes very little when a member is sampled and not replaced. ↩︎