the following shows a corn maze for participants to walk through. the participants can only go forward and when there is a fork in the path you may assume that people are equally likely to take any one of the paths. it costs $5.00 to enter the maze and each participant gets $10.00 if they come out in a prize area.

a person creates a game in which a person pays $5.00 to play. then, they draw a single card from a standard shuffled deck of cards.

if they draw a face card, they win $10.00 (i. e. they get their $5.00 back and get an extra $5.00).

if they draw an ace, they win $20.00 (i. e. they get their $5.00 back and get an extra $15.00).

with any other draw they lose their money.

what is the expected money value playing this game?

-$5.00

-$1.15

$1.31

$0.38

an opaque bag contains 5 green marbles, 3 blue marbles, and 2 red marbles. if two consecutive marbles are drawn without replacement what is the probability that the first one will be green and the second one will be red?

0.51

0.10

0.11

0.16

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.

this is a good question! i’ll do some research and what i find, i’ll contact you asap when i have something! ✊