<>= 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}