Question

A student argues that there are $11$ possible outcomes (sum of numbers on both dice) $2,3,4,5,6,7,8,9,10,11$ and $12$ when we throws two dice simultaneously. Therefore each of them has a probability $\frac{1}{11}.$ Do you agree with this argument? Justify your answer.

Answer 1

Event:Sum of two dice	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
Probability	$\frac{1}{36}$	$\frac{2}{36}=\frac{1}{18}$	$\frac{3}{36}=\frac{1}{12}$	$\frac{4}{36}=\frac{1}{9}$	$\frac{5}{36}$	$\frac{6}{36}=\frac{1}{6}$	$\frac{5}{36}$	$\frac{4}{36}=\frac{1}{9}$	$\frac{3}{36}=\frac{1}{12}$	$\frac{2}{36}=\frac{1}{18}$	$\frac{1}{36}$

Total number of outcomes$=36$

It can be seen that each term does not have a probability of $\frac{1}{11}$

$\therefore$, it is an incorrect argument.