If 20% of the people who shop at a local grocery store buy apples, what is the probability that it will take no more than 5 customers to find one who buys apples? Which simulation design has an appropriate device and a correct trial for this problem? A) Roll a fair die where 1-2 are buying apples and 3-6 are not buying apples. Roll the die until you get a 1 or 2. Record the number of rolls it took you. B) Using a random digits table select one digit numbers where 0-2 is a customer who buys apples and 3-9 is a customer who does not. Keep selecting one digit numbers until you get a 0-2. Record the number of digits selected. C) Using a random digits table select one digit numbers where 0-1 is a customer who buys apples and 2-9 is a customer who does not. Keep selecting one digit numbers until you get a 0 or 1. Record the number of digits selected. D) Spin a spinner that is split up into 5 sections, where 2 sections are a success of buying apples and the other three sections are not buying apples. Keep spinning until you get someone that buys apples. Record the number of spins it took you.

could you provide figure.

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.