Question

The probability of getting a total of atleast $6$ in the simultaneously throw of three dice is

• A. $\frac{6}{108}$
• B. $\frac{5}{27}$
• C. $\frac{1}{24}$
• D. $\frac{103}{108}$

Answer 1

The correct option is D

$\frac{103}{108}$

Explanation of the correct option.

Compute the probability.

If three dice are thrown, total sample space $=216$

The event of getting the sum less than $6$ $=\left\{\left(1,1,1\right),\left(1,1,2\right),\left(1,2,1\right),\left(2,1,1\right),\left(1,2,2\right),\left(2,2,1\right),\left(2,1,2\right),\left(1,1,3\right),\left(3,1,1\right),\left(1,3,1\right)\right\}$

Since the number of outcomes, in which the sum less than $6$ is $10$.

The probability of getting a total less than $6$ $=\frac{10}{216}$

Therefore, the probability of getting a total of atleast $6$ $=1-\frac{10}{216}$

The probability of getting a total of atleast $6$ $=\frac{103}{108}$

Hence option $\left(D\right)$ is the correct option.