Question

The probability of getting a total of at least $6$ in the simultaneously throw of three dice is
A. $\dfrac{6}{{108}}$
B. $\dfrac{5}{{27}}$
C. $\dfrac{1}{{24}}$
D. $\dfrac{{103}}{{108}}$

Answer 1

Hint: In this question we have to determine the probability of the three dice when they are thrown simultaneously, the sum would be at least $6$ hence, first we have to find the total number of outcomes i.e., we throw one dice the outcome will be $6$ and for two dice it will be $6 \times 6$ and for three it will be $6 \times 6 \times 6$.

Formula used:
The basic formula used for evaluating probability here is:
$Probability{\text{ }}of{\text{ }}getting{\text{ }}an{\text{ }}outcome = \dfrac{{Number{\text{ }}of{\text{ }}getting{\text{ }}favourable{\text{ }}outcomes}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}outcomes}}$
and, $Probability{\text{ }}\left( {of{\text{ }}getting{\text{ }}an{\text{ }}event} \right) = 1 - Probability{\text{ }}\left( {of{\text{ }}not\,getting{\text{ }}an{\text{ }}event} \right)$

Complete Step by Step Solution:
Total number of outcomes if three dice thrown simultaneously will be $6 \times 6 \times 6 = 216$.
Now, chance of getting a sum of less than 6 is as follow: -
$\left( {1,1,1} \right),\left( {1,1,2} \right),\left( {1,2,1} \right),\left( {2,1,1} \right),\left( {1,2,2} \right),\left( {2,2,1} \right),\left( {2,1,2} \right),\left( {1,1,3} \right),\left( {3,1,1} \right),\left( {1,3,1} \right)$
Therefore, the total number of outcomes in which the sum is less than $6$are $10$.
Now, we know $P = \dfrac{E}{N}$, where,
P = Probability of getting an outcome
E = Number of getting favorable outcomes
N = Total Number of outcomes
Let us consider the probability of getting a sum at least $6$ is $P(E)$ then, the probability of getting a sum less than $6$will be $P(not\,E)$.
The probability of getting a sum less than $6$,$P(not\,E) = \dfrac{{10}}{{216}}$
Now, by definition of probability we know that $P\left( E \right) = 1 - P\left( {not\,E} \right)$
Therefore, the probability of getting a sum at least $6\,,\,P(E)$
$P(E) = 1 - \dfrac{{10}}{{216}}$
$ \Rightarrow P(E) = \dfrac{{216 - 10}}{{216}} = \dfrac{{206}}{{216}}$
On further calculation, we get
$ \Rightarrow P(E) = \dfrac{{103}}{{108}}$
Thus, the probability of getting a required event is $\dfrac{{103}}{{108}}$.
Hence, the correct option is (D) $\dfrac{{103}}{{108}}$ .

Note: To determine the required probability, first we need to find the favorable outcomes according to the given question; after that use the formula $P = \dfrac{E}{N}$ to determine the opposite of the probability and after that use the formula $P\left( E \right) = 1 - P\left( {\overline E } \right)$ to calculate the required probability.