Answers to exercises

Exercise 1
  1. Let \(X\) be the number of winning wrappers in one week (7 days). Then \(X \stackrel{\mathrm{d}}{=} \mathrm{Bi}(7,\frac{1}{6})\).
  2. \(\Pr(X = 0) = \bigl(\dfrac{5}{6}\bigr)^7 \approx 0.279\).
  3. The purchases are assumed to be independent, and hence the outcome of the first six days is irrelevant. The probability of a winning wrapper on any given day (including on the seventh day) is \(\dfrac{1}{6}\).
  4. Let \(Y\) be the number of winning wrappers in six weeks (42 days). Then \(Y \stackrel{\mathrm{d}}{=} \mathrm{Bi}(42,\frac{1}{6})\). Hence, \begin{align*} \Pr(Y \geq 3) &= 1 - \Pr(Y \leq 2) \\ &= 1 - \bigl[\Pr(Y=0) + \Pr(Y=1) + \Pr(Y=2)\bigr] \\ &= 1 - \bigl[\bigl(\tfrac{5}{6}\bigr)^{42} + 42 \times \tfrac{1}{6} \times \bigl(\tfrac{5}{6}\bigr)^{41} + 861 \times \bigl(\tfrac{1}{6}\bigr)^2 \times \bigl(\tfrac{5}{6}\bigr)^{40}\bigr] \\ &\approx 1 - \bigl[0.0005 + 0.0040 + 0.0163\bigr] \approx 0.979. \end{align*} This could also be calculated with the help of the \(\sf \text{BINOM.DIST}\) function in Excel.
  5. Let \(U\) be the number of winning wrappers in \(n\) purchases. Then \(U \stackrel{\mathrm{d}}{=} \mathrm{Bi}(n,\frac{1}{6})\) and so \begin{align*} \Pr(U \geq 1) \geq 0.95 \ &\iff\ 1 - \Pr(U = 0) \geq 0.95 \\ &\iff\ \Pr(U = 0) \leq 0.05 \\ &\iff\ \bigl(\tfrac{5}{6}\bigr)^n \leq 0.05 \\ &\iff\ n\,\log_e\bigl(\tfrac{5}{6}\bigr) \leq \log_e(0.05) \\ &\iff\ n \geq \dfrac{\log_e(0.05)}{\log_e\bigl(\tfrac{5}{6}\bigr)} \approx 16.431. \end{align*} Hence, he needs to purchase a bar every day for 17 days.
Exercise 2
  1. Binomial.
  2. Not binomial. The weather from one day to the next is not independent. Even though, for example, the chance of rain on a day in March may be 0.1 based on long-term records for the city, the days in a particular fortnight are not independent.
  3. Not binomial. When sampling without replacement from a small population of tickets, the chance of a 'success' (in this case, a blue ticket) changes with each ticket drawn.
  4. Not binomial. There is likely to be clustering within classes; the students within a class are not independent from each other.
  5. Binomial.
Exercise 3

This problem is like that of Casey and the winning wrappers. Let \(X\) be the number of sampled quadrats containing the species of interest. We assume that \(X \stackrel{\mathrm{d}}{=} \mathrm{Bi}(n,k)\). We want \(\Pr(X \geq 1) \geq 0.9\), since we only need to observe the species in one quadrat to know that it is present in the forest. We have

\begin{align*} \Pr(X \geq 1) \geq 0.9 \ &\iff\ \Pr(X = 0) \leq 0.1 \\ &\iff\ (1-k)^n \leq 0.1 \\ &\iff\ n \geq \dfrac{\log_e(0.1)}{\log_e(1-k)}. \end{align*}

For example: if \(k = 0.1\), we need \(n \geq 22\); if \(k=0.05\), we need \(n \geq 45\); if \(k=0.01\), we need \(n \geq 230\).

Exercise 4
  1. \(p\) 0.01 0.05 0.2 0.35 0.5 0.65 0.8 0.95 0.99
    \(\mu_X\) 0.2 1 4 7 10 13 16 19 19.8
    \(\mathrm{sd}(X)\) 0.445 0.975 1.789 2.133 2.236 2.133 1.789 0.975 0.445


  2. The standard deviation is smallest when \(|p-0.5|\) is largest; in this set of distributions, when \(p=0.01\) and \(p = 0.99\). The standard deviation is largest when \(p = 0.5\).
Exercise 5
  1. The graph of the variance of \(X\) as a function of \(p\) is as follows.

    The graph of the variance as a function of p is shown. Var(X) = np(1-p).

    Figure 2: Variance of \(X\) as a function of \(p\), where \(X \stackrel{\mathrm{d}}{=} \mathrm{Bi}(n,p)\).

  2. The variance function is \(f(p) = np(1-p)\). So \(f'(p) = n(1-2p)\). Hence \(f'(p) = 0\) when \(p = 0.5\), and this is clearly a maximum.
  3. When \(p = 0.5\), we have \(\mathrm{var}(X) = \dfrac{n}{4}\) and \(\mathrm{sd}(X) = \dfrac{\sqrt{n}}{2}\).
Exercise 6

Yes, it is true that, as \(n\) tends to infinity, the largest value of \(p_X(x)\) tends to zero. We will not prove this explicitly, but make the following two observations.