What is the total number of outcomes if the four dice are thrown together

When rolling two dice, distinguish between them in some way: a first one and second one, a left and a right, a red and a green, etc. Let (a,b) denote a possible outcome of rolling the two die, with a the number on the top of the first die and b the number on the top of the second die. Note that each of a and b can be any of the integers from 1 through 6. Here is a listing of all the joint possibilities for (a,b):

(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)Note that there are 36 possibilities for (a,b). This total number of possibilities can be obtained from the multiplication principle: there are 6 possibilities for a, and for each outcome for a, there are 6 possibilities for b. So, the total number of joint outcomes (a,b) is 6 times 6 which is 36. The set of all possible outcomes for (a,b) is called the sample space of this probability experiment.

With the sample space now identified, formal probability theory requires that we identify the possible events. These are always subsets of the sample space, and must form a sigma-algebra. In an example such as this, where the sample space is finite because it has only 36 different outcomes, it is perhaps easiest to simply declare ALL subsets of the sample space to be possible events. That will be a sigma-algebra and avoids what might otherwise be an annoying technical difficulty. We make that declaration with this example of two dice.

With the above declaration, the outcomes where the sum of the two dice is equal to 5 form an event. If we call this event E, we have

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

Note that we have listed all the ways a first die and second die add up to 5 when we look at their top faces.

Consider next the probability of E, P(E). Here we need more information. If the two dice are fair and independent , each possibility (a,b) is equally likely. Because there are 36 possibilities in all, and the sum of their probabilities must equal 1, each singleton event {(a,b)} is assigned probability equal to 1/36. Because E is composed of 4 such distinct singleton events, P(E)=4/36= 1/9.

In general, when the two dice are fair and independent, the probability of any event is the number of elements in the event divided by 36.

What if the dice aren't fair, or aren't independent of each other? Then each outcome {(a,b)} is assigned a probability (a number in [0,1]) whose sum over all 36 outcomes is equal to 1. These probabilities aren't all equal, and must be estimated by experiment or inferred from other hypotheses about how the dice are related and and how likely each number is on each of the dice. Then the probability of an event such as E is the sum of the probabilities of the singleton events {(a,b)} that make up E.

Hint: To find the probability we need to find the sample space first. The sample space is nothing but the number of values in whole. The total number of values is called sample space. Then find the value of a number that satisfies the condition. To find the probability we need to divide the numbers that satisfy the condition to the sample space.
Formula:
Sample space be $n(s)$.
The total values satisfy the equation $n(A)$.
$p(A) = \dfrac{{n(A)}}{{n(s)}}$

Complete step-by-step answer:
Given that the four dice are drawn,
If a single dice is drawn, the total number of throws is ${6^1} = 6$.
If four dice are drawn, the total number of throws is ${6^4} = 1296$.
The number of sample spaces is the number of throws.
$n(s) = 1296$.
The condition is that the sum of the numbers appearing on them is $13$.
To find the throws which satisfy the condition, we have to use permutations.
The values on the dice are $1,2,3,4,5$and $6$.
The possible ways of getting $13$as a sum are \[
  (1,1,5,6),(1,2,4,6),(1,3,3,6),(1,2,5,5),(1,3,4,5) \\
  (2,2,6,3),(2,2,5,4),(3,3,2,5),(3,3,3,4),(4,4,4,1),(4,4,3,2) \\
 \]
We have to find each permutations,
Permutation of $(4,4,3,2)$
$
  P(4,4,3,2) = \dfrac{{4!}}{{2!}} \\
  P(4,4,3,2) = 12 \\
 $
Permutation of $(4,4,4,1)$
$
  P(4,4,4,1) = \dfrac{{4!}}{{3!}} \\
  P(4,4,4,1) = 4 \\
 $
Permutation of $(3,3,3,4)$
$
  P(3,3,3,4) = \dfrac{{4!}}{{3!}} \\
  P(3,3,3,4) = 4 \\
 $
Permutation of $(3,3,2,5)$
$
  P(3,3,2,5) = \dfrac{{4!}}{{2!}} \\
  P(3,3,2,5) = 12 \\
 $
Permutation of $(2,2,5,4)$
$
  P(2,2,5,4) = \dfrac{{4!}}{{2!}} \\
  P(2,2,5,4) = 12 \\
 $
Permutation of $(2,2,3,6)$
$
  P(2,2,3,6) = \dfrac{{4!}}{{2!}} \\
  P(2,2,3,6) = 12 \\
 $
Permutation of $(1,1,5,6)$
$
  P(1,1,5,6) = \dfrac{{4!}}{{2!}} \\
  P(1,1,5,6) = 12 \\
 $
Permutation of $(1,2,4,6)$
$
  P(1,2,4,6) = 4! \\
  P(1,1,5,6) = 24 \\
 $
Permutation of $(1,3,3,6)$
$
  P(1,3,3,6) = \dfrac{{4!}}{{2!}} \\
  P(1,3,3,6) = 12 \\
 $
Permutation of $(1,2,5,5)$
$
  P(1,2,5,5) = \dfrac{{4!}}{{2!}} \\
  P(1,2,5,5) = 12 \\
 $
Permutation of $(1,3,5,4)$
$
  P((1,3,5,4)) = 4! \\
  P((1,3,5,4)) = 24 \\
 $
By adding all the permutation, we will get to know the number of events that satisfy the condition, such as
$n(A) = 12 + 24 + 12 + 12 + 24 + 12 + 12 + 12 + 4 + 4 + 12$
By adding the terms in the above equation, we will get
$n(A) = 140$
As we know the formula for probability $p(A) = \dfrac{{n(A)}}{{n(s)}}$.
Substituting $n(A) = 140$and $n(s) = 1296$,
$p(A) = \dfrac{{140}}{{1296}}$
By dividing the numerator and the denominator to an extent, we get
$p(A) = \dfrac{{35}}{{324}}$
The probability that the sum of the numbers appearing on them is $13$, is $\dfrac{{32}}{{324}}$.

Note: Remember to find the sample space first be careful while selecting it because there would be differences. Careful while taking values that satisfy the equation. The formula for the finding probability is the number of values that satisfy the condition to the number of sample spaces.
Divide the probability to the maximum extent to get the correct final answer

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