Solution 7.7.3.1.

  1. One. The first meal will certainly have a book that you have not received yet.
  2. This is a geometric distribution with p=4/5. P(X > 4) = 1 - F(4) = \((1 - 4/5)^4 = \frac{1}{625}\) which is very small. Note, in this case you would have needed to receive the same title randomly for all of the first four kids meal purchases. If this were to ever happen, please let the people at the counter know and it will be their pleasure to swap out for a new title.
  3. Use the geometric distribution five times with changing values for p. For the first book p = 1 means you are certain to get a new title. For the second book title the probability of success is p=4/5; for the third book title the probability of success is p=3/5; for the fourth the probability is p=2/5; and for the last the probability is p = 1/5. Using the mean as 1/p in each case and accumulating these gives the total expected number of meals to purchase as
    \begin{align*} & 1 + \frac{1}{\frac{4}{5}} + \frac{1}{\frac{3}{5}} + \frac{1}{\frac{2}{5}} + \frac{1}{\frac{1}{5}} \\ & = 1 + \frac{5}{4} + \frac{5}{3} + \frac{5}{2} + \frac{5}{1} \\ & = \frac{12 + 15 + 25 + 30 + 60}{12} \\ & = \frac{142}{12} = 11.833 \end{align*}
    and so you would need 12 kids meals. If this were to happen, please be certain to donate the "extra" books to an organization that works with kids or directly to some kids that you might know.
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