Question

A student argues that ‘there are 11 possible outcomes $2,3,4,5,6,7,8,9,10,11and12.$ Therefore, each of them has a probability $\frac{1}{11}$$$\begin{gathered} P(sum10)=\frac{3}{36} \\ E(sum11)=(5,6),and(6,5) \end{gathered}$$

Do you agree with this argument? Justify your Solution.

Answer 1

If $2$ dice are thrown, the possible events are:

$$\begin{gathered} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{gathered}$$

So, the total number of events: $6\times 6=36$

(i) It is given that to get the sum as $2,$ the probability is $\frac{1}{36}$ as the only possible outcomes $=(1,1)$

For getting the sum as $3,$ the possible events (or outcomes) $=E(sum3)=(1,2)and(2,1)$

So, $P(sum3)=2/36$

Similarly,

$E(sum4)=(1,3),(3,1),and(2,2)$

So,

$$\begin{gathered} P(sum4)=\frac{3}{36} \\ E(sum5)=(1,4),(4,1),(2,3),and(3,2) \end{gathered}$$

So,

$$\begin{gathered} P(sum5)=\frac{4}{36} \\ E(sum6)=(1,5),(5,1),(2,4),(4,2),and(3,3) \end{gathered}$$

So,

$$\begin{gathered} P(sum6)=\frac{5}{36} \\ E(sum7)=(1,6),(6,1),(5,2),(2,5),(4,3),and(3,4) \end{gathered}$$

So,

$$\begin{gathered} P(sum7)=\frac{6}{36} \\ E(sum8)=(2,6),(6,2),(3,5),(5,3),and(4,4) \end{gathered}$$

So,

$$\begin{gathered} P(sum8)=\frac{5}{36} \\ E(sum9)=(3,6),(6,3),(4,5),and(5,4) \end{gathered}$$

So,

$$\begin{gathered} P(sum9)=\frac{4}{36} \\ E(sum10)=(4,6),(6,4),and(5,5) \end{gathered}$$

So,

$$\begin{gathered} P(sum10)=\frac{3}{36} \\ E(sum11)=(5,6),and(6,5) \end{gathered}$$

So,

$$\begin{gathered} P(sum11)=\frac{2}{36} \\ E(sum12)=(6,6) \end{gathered}$$

So, $P(sum12)=\frac{1}{36}$

So, the table will be as:

Event: The sum on 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) The argument is not correct as it is already justified in (i) that the number of all possible outcomes is $36andnot11$.