Question

If four dice are thrown together. The probability that the sum of the number appearing on them is $13$, is

Answer 1

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