<>=
sc <- NULL
while(is.null(sc)) {
## success probability in percent (= pay with card)
p <- sample(15:30, size = 1)
## number of attempts (= customers in queue)
n <- sample(6:9, size = 1)
## minimum number of successes (= customers who pay with card)
k <- sample(1:3, 1)
## compute the correct solution in percent
sol <- 100 * pbinom(k - 1, size = n, prob = p/100, lower.tail = FALSE)
## use one of two typical errors: 1-p vs. p, pbinom vs. dbinom
err1 <- 100 * pbinom(k - 1, size = n, prob = 1 - p/100, lower.tail = FALSE)
err2 <- 100 * dbinom(k, size = n, prob = p/100)
err <- sample(c(err1, err2), 1)
sc <- num_to_schoice(sol, wrong = err, range = c(2, 98), delta = 0.1)
}
@
\begin{question}
According to a recent survey $\Sexpr{100 - p}$ percent of all customers
in grocery stores pay cash while the rest use their credit or cash card. You
are currently waiting in the queue at the checkout of a grocery story with
$\Sexpr{n}$ customers in front of you.
What is the probability (in percent) that $\Sexpr{k}$ or more of the
other customers pay with their card?
<>=
answerlist(sc$questions)
@
\end{question}
\extype{schoice}
\exsolution{\Sexpr{mchoice2string(sc$solutions)}}
\exname{binomial v4}