Question

(i) Complete the following table:

Event: 'Sum on $2$ dice'	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
Probability	$\frac{1}{36}$						$\frac{5}{36}$				$\frac{1}{36}$

(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

Event: Sum of 2 dice	2	3	4	5	6	7	8	9	10	11	12
Probability	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

(i) From the table it can be observed that,

To get sum as 2, possible outcomes $=(1,1)$

To get sum as 3, possible outcomes $=(2,1),(1,2)$

To get sum as 4, possible outcomes $=(3,1),(1,3),(2,2)$

To get sum as 5, possible outcomes $=(2,3),(3,2),(1,4),(4,1)$

To get sum as 6, possible outcomes $=(1,5),(5,1),(2,4),(4,2),(3,3)$

To get sum as 7, possible outcomes $=(1,6),(6,7),(3,4),(4,3),(2,5),(5,2)$

To get sum as 8, possible outcome $=(2,6),(6,2),(3,5),(5,3),(4,4)$

To get sum as 9, possible outcomes $=(3,6),(6,3),(4,5),(5,4)$

To get sum as 10, possible outcome $=(4,6),(6,4),(5,5)$

To get sum as 11, possible outcome $=(5,6),(6,5)$

To get sum as 12, possible outcome $=(6,6)$

(ii) The probability of each of these sums will not be $\frac{1}{11}$, as these sums are not equally likely.