Question

(i) Complete the following table:

$\begin{array}{cc}\text{Event:'Sum of 2 dice'}& \text{Probability}\\ 2& \frac{1}{36}\\ 3& \\ 4& \\ 5& \\ 6& \\ 7& \\ 8& \frac{5}{36}\\ 9& \\ 10& \\ 11& \\ 12& \frac{1}{36}\end{array}$

(ii) A 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 $\frac{1}{11}$. Do you agree with this argument? Justify your answer.

Answer 1

Elementary events associated to the random experiment of throwing two dice 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)

Therefore,total number of elementary events =$6\times 6=36$

Let A be the event of getting the sum as 3.

The elementary events favourable to event A are (1, 2) and (2, 1)

Clearly, favourable number of elementary events = 2

Hence, required probability = $\frac{2}{36}$

Let A be the event of getting the sum as 4.

The elementary events favourable to event A are (1, 3), (3, 1) and (2, 2)

Clearly, favourable number of elementary events = 3

Hence, required probability = $\frac{3}{36}$

Let A be the event of getting the sum as 5.

The elementary events favourable to event A are (1, 4), (4, 1), (2, 3) and (3, 4)

Clearly, favourable number of elementary events = 4

Hence, required probability = $\frac{4}{36}=\frac{1}{9}$

Let A be the event of getting the sum as 6.

The elementary events favourable to event A are (1, 5), (5,1), (2, 4), (4, 2) and (3,3)

Clearly, favourable number of elementary events = 5

Hence, required probability = $\frac{5}{36}$

Let A be the event of getting the sum as 7.

The elementary events favourable to event A are (1, 6), (6, 1), (2, 5), (5, 2), (3,4), (4, 3)

Clearly, favourable number of elementary events = 6

Hence, required probability = $\frac{6}{36}=\frac{1}{6}$

Let A be the event of getting the sum as 9.

The elementary events favourable to event A are (3, 6), (6, 3), (4, 5) and (5, 4)

Clearly, favourable number of elementary events = 4

Hence, required probability = $\frac{4}{36}=\frac{1}{9}$

Let A be the event of getting the sum as 10.

The elementary events favourable to event A are (4, 6), (6, 4), (5, 5)

Clearly, favourable number of elementary event = 3

Hence, required probability = $\frac{3}{36}=\frac{1}{12}$

Let A be the event of getting the sum as 11.

The elementary events favourable to event A are (5, 6), (6, 5)

Clearly, favourable number of elementary events = 2

Hence, required probability = $\frac{2}{36}=\frac{1}{18}$

Thus, the complete table is as below:

i)

$\begin{array}{cc}\text{Event:'Sum of 2 dice'}& \text{Probability}\\ 2& \frac{1}{36}\\ 3& \frac{2}{36}\\ 4& \frac{3}{36}\\ 5& \frac{4}{36}\\ 6& \frac{5}{36}\\ 7& \frac{6}{36}\\ 8& \frac{5}{36}\\ 9& \frac{4}{36}\\ 10& \frac{3}{36}\\ 11& \frac{2}{36}\\ 12& \frac{1}{36}\end{array}$

ii) No. Justification has already been given in part (i).