Question 8.6: Marco’s company organizes a raffle at an end-of-year functio......
Marco’s company organizes a raffle at an end-of-year function. There are 4000 raffle tickets to be sold, of which 500 win a prize. The price of each ticket is €1.50. The value of the prizes, which are mostly electrical appliances produced by the company, varies between €80 and €250, with an average value of €142.
(a) Marco wants to have a 99 % guarantee of receiving three prizes. How much money does he need to spend? Use R to solve the question.
(b) Use R to plot the function which describes the relationship between the number of tickets bought and the probability of winning at least three prizes.
(c) Given the value of the prizes and the costs of the tickets, is it worth taking part in the raffle?
(a) Let X be a random variable describing “the number x of winning tickets among n bought tickets”; then X follows the hypergeometric distribution X ∼ H(n, 500, 4000). We need to determine n for the conditions specified.
We are interested in
P(X ≥ 3) = 1 − P(X = 2) − P(X = 1) − P(X = 0).
Using the PMF of the hypergeometric distribution
P(X = x)= \frac{\left(\begin{matrix} M \\ x \end{matrix} \right) \left(\begin{matrix} N-M \\ n-x \end{matrix} \right)}{\left(\begin{matrix} N \\ n \end{matrix} \right)} ,
this equates to
P(X ≥ 3) = 1 − \frac{\left(\begin{matrix} 500 \\ 2 \end{matrix} \right) \left(\begin{matrix} 4000-500 \\ n-2 \end{matrix} \right)}{\left(\begin{matrix} 4000 \\ n \end{matrix} \right)}-\frac{\left(\begin{matrix} 500 \\ 1 \end{matrix} \right) \left(\begin{matrix} 4000-500 \\ n-1 \end{matrix} \right)}{\left(\begin{matrix} 4000 \\ 1 \end{matrix} \right)}-\frac{\left(\begin{matrix} 500 \\ 0 \end{matrix} \right) \left(\begin{matrix} 4000-500 \\ n \end{matrix} \right)}{\left(\begin{matrix} 4000 \\ n \end{matrix} \right)}.
We have the following requirement:
1 − \frac{\left(\begin{matrix} 500 \\ 2 \end{matrix} \right) \left(\begin{matrix} 4000-500 \\ n-2 \end{matrix} \right)}{\left(\begin{matrix} 4000 \\ n \end{matrix} \right)}-\frac{\left(\begin{matrix} 500 \\ 1 \end{matrix} \right) \left(\begin{matrix} 4000-500 \\ n-1 \end{matrix} \right)}{\left(\begin{matrix} 4000 \\ 1 \end{matrix} \right)}-\frac{\left(\begin{matrix} 500 \\ 0 \end{matrix} \right) \left(\begin{matrix} 4000-500 \\ n \end{matrix} \right)}{\left(\begin{matrix} 4000 \\ n \end{matrix} \right)}\overset{!}{\geq } 0.99.
To solve this equation, we can program this function for P(X > 3; n) in R and evaluate it for different numbers of tickets sold, e.g. between 50 and 100 tickets:
| raffle <- function(n){ p <- 1-((choose(500,2)*choose(3500,n-2))/(choose(4000,n))) -((choose(500,1)*choose(3500,n-1))/(choose(4000,n))) -((choose(500,0)*choose(3500,n))/(choose(4000,n))) return(p) } raffle(50:100) raffle(63:64) |
The output shows that at least 64 tickets need to be bought to have a 99% guarantee that at least three tickets win. This equates to spending € 96.
(b) We can plot the function as follows:
| nb <- seq(1:75) plot(nb,tombola(nb),type=’l’) |
Figure B.16 shows the relationship between the number of tickets bought and the probability of having at least three winning tickets.
(c) The solution of (a) shows that it is well worth taking part in the raffle: Marco pays €96 and with a probability of 99 % and he wins at least three prizes which are worth €142 · 3 = 426. More generally, the money generated by the raffle is €1.50 × 4000 = 6000, but the prizes are worth €142 · 500 = 71, 000. One may suspect that the company produces the appliances much more cheaply than they are sold for and is thus so generous.
