Question

A die is numbered in such a way that its faces show the numbers $1,{\text{ }}2,{\text{ }}2,{\text{ }}3,{\text{ }}3,{\text{ }}6$. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:
What is the probability that the total score is
(i). even?
(ii). $6$?
(iii). at least $6$?

+	1	2	2	3	3	6
1	2	3	3	4	4	7
2	3	4	4	5	5	8
2					5
3
3			5			9
6	7	8	8	9	9	12

Answer 1

We will complete the table first. Then using the data from the table we can find probability for the given requirement. We have to use the probability formula to find the probability of the given requirement.
Probability is a type of ratio where we compare how many times an outcome can occur compared to all possible outcomes.
${\text{Probability = }}\dfrac{{\left( {{\text{The number of wanted outcomes}}} \right)}}{{\left( {{\text{The total possible outcomes}}} \right)}}$

Complete step-by-step answer:
It is given that, the die is numbered in such a way that its faces show the numbers $1,{\text{ }}2,{\text{ }}2,{\text{ }}3,{\text{ }}3,{\text{ }}6$. It is thrown two times and the total score in two throws is noted.
We need to complete the table and determine the probability left blank in the table.
Adding the face values of number in first throw and number in second throw to complete the table.
For face $1$ of number in first throw and face $2$ of number in second throw $= 1 + 2 = 3$
For face $2$ of number in first throw and face $2$ of number in second throw $= 2 + 2 = 4$
For face $2$ of number in first throw and face $2$ of number in second throw $= 2 + 2 = 4$
For face $3$ of number in first throw and face $2$ of number in second throw $= 3 + 2 = 5$
For face $6$ of number in first throw and face $2$ of number in second throw $= 6 + 2 = 8$
For face $1$ of number in first throw and face $3$ of number in second throw $= 1 + 3 = 4$
For face $2$ of number in first throw and face $3$ of number in second throw $= 2 + 3 = 5$
For face $2$ of number in first throw and face $3$ of number in second throw $= 2 + 3 = 5$
For face $3$ of number in first throw and face $3$ of number in second throw $= 3 + 3 = 6$
For face $3$ of number in first throw and face $3$ of number in second throw $= 3 + 3 = 6$
For face $6$ of number in first throw and face $3$ of number in second throw $= 6 + 3 = 9$
For face $1$ of number in first throw and face $3$ of number in second throw $= 1 + 3 = 4$
For face $2$ of number in first throw and face $3$ of number in second throw $= 2 + 3 = 5$
For face $3$ of number in first throw and face $3$ of number in second throw $= 3 + 3 = 6$
For face $3$ of number in first throw and face $3$ of number in second throw $= 3 + 3 = 6$
Hence the completed table,

+	1	2	2	3	3	6
1	2	3	3	4	4	7
2	3	4	4	5	5	8
2	3	4	4	5	5	8
3	4	5	5	6	6	9
3	4	5	5	6	6	9
6	7	8	8	9	9	12

The total number of possibilities when two dice are thrown is $36$.
The possibilities are

$$(1,1),(1,2),(1,2),(1,3),(1,3),(1,6) \\\ (2,1),(2,2),(2,2),(2,3),(2,3),(2,6) \\\ (2,1),(2,2),(2,2),(2,3),(2,3),(2,6) \\\ (3,1),(3,2),(3,2),(3,3),(3,3),(3,6) \\\ (3,1),(3,2),(3,2),(3,3),(3,3),(3,6) \\\ (6,1),(6,2),(6,2),(6,3),(6,3),(6,6) \\\$$

So the table can be completed.
So the number of possible outcomes is $36$.
(i). When the total score is even the possible cases are $2,{\text{ }}4,{\text{ }}6,{\text{ }}8,\,\,12$
Now we get from the table, the number of favorable outcomes$= 18$
Thus the probability of the total score is even is $\dfrac{{18}}{{36}} = \dfrac{1}{2}$ .
(ii). When the total score is $6$,
We get from the table, the number of favorable outcomes $= 4$
Thus the probability of the total score $6$ is $\dfrac{4}{{36}} = \dfrac{1}{9}$ .
(iii). When the total score is at least $6$,
At least $6$ means the total score is either $6$ or greater than $6$.
So the Score can be $6,{\text{ }}7,{\text{ }}8,\,{\text{ }}9,{\text{ }}12$
We get from the table, the number of favorable outcomes $= 15$
Thus the probability of the total score is at least $6$ is $\dfrac{{15}}{{36}} = \dfrac{5}{{12}}$

Note: The problem is easy to solve. But we have to concentrate on completion of the table. Because we may go wrong in calculating the sum the two throws. We find the probability of the given requirements using the completed table. Therefore, we give most importance to completion of the table.