Let <span>Jacob, Carol, Geraldo, Meg, Earvin, Dora, Adam, and Sally be represented by the letters J, C, G, M, E, D, A, and S respectively. </span>
<span>In part IV we are asked:
</span><span>What is the sample space of the pairs of potential clients that could be chosen?
</span><span>
Since the Sample Space is the set of all possible outcomes, we need to make a set (a list) of all the possible pairs, which are as follows:
{(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S)
, </span>(C, G), (C, M), (C, E), (C, D), (C, A), (C, S)
<span>
</span> , (G, M), (G, E), (G, D), (G, A), (G, S)
<span>
,</span>(M, E), (M, D), (M, A), (M, S)
<span>
, </span>(E, D), (E, A), (E, S) <span>
, </span>(D, A), (D, S)
, (A, S).}
We can check that the number of the elements of the sample space, n(S) is
1+2+3+4+5+6+7=28.
This gives us the answer to the first question: <span>How many pairs of potential clients can be randomly chosen from the pool of eight candidates?
(Answer: 28.)
II) </span><span>What is the probability of any particular pair being chosen?
</span>
The probability of a particular pair to be picked is 1/28, as there is only one way of choosing a particular pair, out of 28 possible pairs.
III) <span>What is the probability that the pair chosen is Jacob and Meg or Geraldo and Sally?
The probability of choosing (J, M) or (G, S) is 2 out of 28, that is 1/14.
Answers:
I) 28
II) 1/28</span>≈0.0357
III) 1/14≈0.0714
IV)
{(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S)
, (C, G), (C, M), (C, E), (C, D), (C, A), (C, S)
, (G, M), (G, E), (G, D), (G, A), (G, S)
,(M, E), (M, D), (M, A), (M, S)
, (E, D), (E, A), (E, S)
, (D, A), (D, S)
, (A, S).}
Answer: Transitive property.
Step-by-step explanation:
First, for the equality we have:
Reflexive:
For all real numbers x, x = x.
Symmetric:
For all real numbers x, y
if x= y, then y = x.
Transitive:
For reals x, y and z.
if x = y, and y = z, then x = z.
Now, let's talk about inequalities.
first, the reflexive property will say that:
x > x.
This has no sense, so this property does not work for inequalities.
Now, the reflexive.
If x > y, then y > x.
Again, this has no sense, if x is larger than y, then we can never have that y is larger than x. This property does not work for inequalities.
Not, the transitive property.
if x > y, and y > z, then x > z.
This is true.
x is bigger than y, and y is bigger than z, then x should also be bigger than z.
x > y > z.
And this also works for the inverse case:
x < y and y < z, then x < z.
So the correct option is transitive property.
