Question

Three dice are thrown simultaneously. The probability of getting the sum greater than 3 and less than 6 is

• A.
  $\frac{3}{34}$
• B.
  $\frac{1}{72}$
• C.
  $\frac{7}{108}$
• D.
  $\frac{1}{24}$

Answer 1

The correct option is D

$\frac{1}{24}$
When three dice are thrown simultaneously, the total number of possible outcomes is $6^3=216.$
The outcomes favourable for getting sum greater than 3 and less than 6 are $(1,1,2),(1,2,1),(2,1,1),(1,2,2),(2,1,2),(2,2,1),(1,1,3),(1,3,1),(3,1,1).$

$\therefore$ Number of favourable outcomes = 9
$\therefore$ Required probability

$=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{9}{216}=\frac{1}{24}$

Hence, the correct answer is option d.