Question

(i). Complete the following table:

Event: Sum of 2 dice	2	3	4	5	6	7	8	9	10	11	12
Probability	1/36						5/36				1/36

(ii). A student argues that there are $11$ possible outcomes $2,3,4,5,6,7,8,9,10,11,12.$Therefore, each of them has a probability $\dfrac{1}{{11}}$.Do you agree with this argument?

Answer 1

We have to complete the given table by using possible outcomes for each value and have to justify whether the statement of student’s is right or wrong.
Probability is a type of ratio where we compare how many times an outcome can occur compared to all possible outcomes.
${\text{Probability = }}\dfrac{{\left( {{\text{The number of wanted outcomes}}} \right)}}{{\left( {{\text{The number of possible outcomes}}} \right)}}$

Complete step-by-step answer:
We need to complete the table and determine the probability left blank in the table.
The total number of possibilities when two dice are thrown is $36$.
The possibilities are

$$(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) \\\$$

When two dice are thrown the possibilities of the sum of the dices are
$2,3,4,5,6,7,8,9,10,11,12.$
To get the sum of the dices as $2$, the possible outcomes =$(1,1)$
To get the sum of the dices as $3$, the possible outcomes =$(1,2),(2,1)$
To get the sum of the dices as $4$, the possible outcomes =$(1,3),(2,2),(3,1)$
To get the sum of the dices as $5$, the possible outcomes =$(1,4),(2,3),(3,2),(4,1)$
To get the sum of the dices as $6$, the possible outcomes =$(1,5),(2,4),(3,3),(4,2),(5,1)$
To get the sum of the dices as $7$, the possible outcomes =$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$
To get the sum of the dices as $8$, the possible outcomes =$(2,6),(3,5),(4,4),(5,3),(6,2)$
To get the sum of the dices as $9$, the possible outcomes =$(3,6),(4,5),(5,4),(6,3)$
To get the sum of the dices as $10$, the possible outcomes =$(4,6),(5,5),(6,4)$
To get the sum of the dices as $11$, the possible outcomes =$(5,6),(6,5)$
To get the sum of the dices as $12$, the possible outcomes =$(6,6)$
Thus the probability of getting the possibilities of the sum of the dices $2,3,4,5,6,7,8,9,10,11,12.$ are $\dfrac{1}{{36}},\dfrac{2}{{36}},\dfrac{3}{{36}},\dfrac{4}{{36}},\dfrac{5}{{36}},\dfrac{6}{{36}},\dfrac{5}{{36}},\dfrac{4}{{36}},\dfrac{3}{{36}},\dfrac{2}{{36}},\dfrac{1}{{36}}$ respectively.
(Since the total number of possibilities when two dice are thrown is $36$.)
So the table can be completed.

Event: Sum of 2 dice	2	3	4	5	6	7	8	9	10	11	12
Probability	1/36	2/36	3/36	4/36	5/36	6/36	5/36	4/36	3/36	2/36	1/36

(ii). A student argues that there are $11$ possible outcomes $2,3,4,5,6,7,8,9,10,11,12.$ Therefore, each of them has a probability $\dfrac{1}{{11}}$.
The student’s argument is wrong, that each of the possible outcomes has a probability $\dfrac{1}{{11}}$ as the probabilities are different for each one as shown above.
The student’s argues that there are $11$ possible outcomes $2,3,4,5,6,7,8,9,10,11,12.$
$\therefore$each of them has a probability $\dfrac{1}{{11}}$. But the probability of possible outcomes for all possible outcomes does not have a probability of $\dfrac{1}{{11}}$.
i.e. It can be seen that each term does not have a probability of $\dfrac{1}{{11}}$. Hence the student’s argument is wrong.

Note: In this problem we may be wrong on the second section to give the argument because the student’s argument is logically correct but it is not correct when it is mathematically.