Question

When a pair of six faced fair dice are thrown, the probability that the sum of the numbers on the two dice is greater than $7$, is

• A. $\frac{1}{3}$
• B. $\frac{5}{12}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$\frac{5}{12}$

The table below shows the numbers that can appear on the two die and the corresponding position in the table shows the sum of the numbers.

Throw	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$2$	$3$	$4$	$5$	$6$	$7$
$2$	$3$	$4$	$5$	$6$	$7$	$8$
$3$	$4$	$5$	$6$	$7$	$8$	$9$
$4$	$5$	$6$	$7$	$8$	$9$	$10$
$5$	$6$	$7$	$8$	$9$	$10$	$11$
$6$	$7$	$8$	$9$	$10$	$11$	$12$

As can be seen from the above table, number of throws in which sum equal to $8$ is $5$, sum equal to $9$ is $4$, sum equal to $10$ is $3$, sum equal to $11$ is $2$, sum equal to $12$ is $1$.

Hence,

Probability $=\frac{\text{Number of throws with sum greater than}7}{\text{Total number of throws}}=\frac{5+4+3+2+1}{36}=\frac{15}{36}=\frac{5}{12}$