In each of the following situations state whether or not the given assignment of probabilities to individual outcomes is legitimate, that is, satisfies the rules of probability. If not, give a reason a) roll a die and record the count of spots on the up-face
P(1)=0 P(2)=1/6 P(3)=1/3 P(4)=1/3 P(5)=1/6 P(6)=0
b) Deal a card from a shuffled deck:
P(clubs)=12/52 P(diamonds)=12/52 P(hearts)= 12/52 P(spades)=16/52
c) Choose a college student at random and record sex and enrollement status
P(female full time)=0.56 P(male full time)=0.44P(female part time)=0.24 P(male part time)= 0.17

Step-by-step explanation:

In each of the situations, probabilities have been assigned to the outcomes of the experiment. Some are legitimate (they correctly satisfy the rules of probability) while others are not.

(A) We know that a die has 6 faces, each with a dot or spot representing numbers 1 to 6. The probabilities assigned to the outcomes 1, 3, 4, and 6 are WRONG. They do not satisfy the law of probability. Each outcome has an equal probability or chance of occurring. There are 6 possible outcomes and each stands a 1/6 chance of occurring when the die is rolled. Hence the probability values assigned to the occurrence of 2 and the occurrence of 5 are correct!

(B) We know the standard French playing-cards has 52 cards in a deck or pack. There are 13 clubs, 13 hearts, 13 diamonds and 13 spades, altogether 52 cards. So in a shuffled deck of cards, the probability that a dealt card will be clubs is 13/52. The probability that it will be hearts is 13/52. The probability that it will be diamonds is 13/52 and the probability that it will be spades is 13/52! Add up the probabilities and they will equal 1.

So the probabilities assigned to these 4 possible outcomes in this question are WRONG.

(C) Now, there is no indication of the total number of students in this college. We need figures on the

- number of males

- number of females

- number of part time students

- number of full time students

Without this information, we cannot ascertain the probabilities required.

Hence, the probability assignments to the 4 groups of college students are WRONG.

wholesale price is the initial price before markup. retail price comes after the markup.

a price of 8 bucks that's marked up by 100%   is

formula: initial price + (percent markup in fraction/decimal)(initial price)

8+(1)(8)

8+8

16

retail price is $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.