Consider an experiment where two 6-sided dice are rolled. We can describe the ordered sample space as below where the first coordinate of the ordered pair represents the first die and the second coordinate represents the second die. If the dice are ‘fair’, then all 36 of these possible outcomes are equally likely. 28. Describe the event E that the sum of the two dice is 5. 29. Find P(E). 30. Let F be the event that the number of the first die is exactly 1 more than the number on the second die. Find P(F|E)

Step-by-step explanation:

Let's start writing the sample space for this experiment :

{ (1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) , (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) , (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) , (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) , (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) , (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6) }

Let's also define the event ⇒

: '' The sum of the two dice is 5 ''

We can describe the event by listing all the favorables cases from ⇒

= { (4,1) , (3,2) , (2,3) , (1,4) }

In order to calculate we are going to divide all the cases favorables to over the total cases from . We can do this because all 36 of these possible outcomes from are equally likely. ⇒

⇒

Finally we are going to define the event ⇒

: '' The number of the first die is exactly 1 more than the number on the second die ''

⇒

= { (2,1) , (3,2) , (4,3) , (5,4) , (6,5) }

Now given two events A and B ⇒

P ( A ∩ B ) =

We define the conditional probability as

with

We need to find therefore we can apply the conditional probability equation :

  (I)

We calculate at the beginning of the question. We only need .

Looking at the sets and we find that (3,2) is the unique result which is in both sets. Therefore is 1 result over the 36 possible results. ⇒

Replacing both probabilities calculated in (I) :

We find out that

step-by-step explanation:

b and c

when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range: