Question

Two dice, one blue and one grey are thrown at the same time. Complete the following table:

Event: Sum on two dice	2	3	4	5	6	7	8	9	10	11	12
probability

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

Answer 1

Hint: In this particular type of question use the concept that when we throw two dice the total number of possible outcomes are ($6 \times 6 = 36$) then calculate individual favorable number of outcomes of the sum of digits such that the sum is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
As we all know that in a dice there are six numbers which are given as, 1, 2, 3, 4, 5, and 6.
So when we throw two dice the total number of possible outcomes are ($6 \times 6 = 36$).
Now the number of ways that the sum of digits is 2 = (1, 1) = 1 way.
Now the number of ways that the sum of digits is 3 = (1, 2), (2, 1) = 2ways.
Now the number of ways that the sum of digits is 4 = (1, 3), (2, 2) and (3, 1) = 3 ways.
Now the number of ways that the sum of digits is 5 = (1, 4), (2, 3), (3, 2) and (4, 1) = 4ways.
Now the number of ways that the sum of digits is 6 = (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1) = 5ways.
Now the number of ways that the sum of digits is 7 = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) = 6ways.
Now the number of ways that the sum of digits is 8 = (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2) = 5ways.
Now the number of ways that the sum of digits is 9 = (3, 6), (4, 5), (5, 4) and (6, 3) = 4ways.
Now the number of ways that the sum of digits is 10 = (4, 6), (5, 5), (6, 4) = 3ways.
Now the number of ways that the sum of digits is 11 = (5, 6), and (6, 5) = 2ways.
Now the number of ways that the sum of digits is 12 = (6, 6) = 1way.
Now as we know that the probability (P) is the ratio of the favorable number of outcomes to the total number of outcomes.
$ \Rightarrow P = \dfrac{{{\text{favorable number of outcomes}}}}{{{\text{total number of outcomes}}}}$
Now the probability that the sum on the dice is two = ($\dfrac{1}{{36}}$).
Now the probability that the sum on the dice is three = ($\dfrac{2}{{36}}$) = ($\dfrac{1}{{18}}$).
Now the probability that the sum on the dice is four = ($\dfrac{3}{{36}}$) = ($\dfrac{1}{{12}}$).
Now the probability that the sum on the dice is five = ($\dfrac{4}{{36}}$) = ($\dfrac{1}{9}$).
Now the probability that the sum on the dice is six = ($\dfrac{5}{{36}}$).
Now the probability that the sum on the dice is seven = ($\dfrac{6}{{36}}$) = ($\dfrac{1}{6}$).
Now the probability that the sum on the dice is eight = ($\dfrac{5}{{36}}$).
Now the probability that the sum on the dice is nine = ($\dfrac{4}{{36}}$) = ($\dfrac{1}{9}$).
Now the probability that the sum on the dice is ten = ($\dfrac{3}{{36}}$) = ($\dfrac{1}{{12}}$).
Now the probability that the sum on the dice is eleven = ($\dfrac{2}{{36}}$) = ($\dfrac{1}{{18}}$).
Now the probability that the sum on the dice is twelve = ($\dfrac{1}{{36}}$).
So the required table become:

Event: Sum on two dice	2	3	4	5	6	7	8	9	10	11	12
probability	$\dfrac{1}{{36}}$	$\dfrac{1}{{18}}$	$\dfrac{1}{{12}}$	$\dfrac{1}{9}$	$\dfrac{5}{{36}}$	$\dfrac{1}{6}$	$\dfrac{5}{{36}}$	$\dfrac{1}{9}$	$\dfrac{1}{{12}}$	$\dfrac{1}{{18}}$	$\dfrac{1}{{36}}$

The student argument is false as from the two dice there are 11 sums possible but due to permutations there are a total 36 outcomes possible because (a, b) is a combination and (b, a) is its permutation, but the sum (a + b) remains the same.
So probability can never be ($\dfrac{1}{{11}}$).
Therefore the student argument is false.
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the probability formula which is stated above, so first find out the total number of outcomes then find out the favorable number of outcomes for all the given sum on two dice then divide them according to probability formula we will get the required answer.