Answer
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Hint: 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.
(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.
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.
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