Question

A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is
A. $\dfrac{1}{729}$
B. $\dfrac{8}{9}$
C. $\dfrac{8}{729}$
D. $\dfrac{8}{243}$

Answer 1

We first explain the concept of empirical probability and how the events are considered. We take the given event of throwing a pair of fair dice thrice and find the number of outcomes. Using the probability theorem of $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( U \right)}$, we get the empirical probability of the events and solve the equation.

Complete answer:
Empirical probability uses the number of occurrences of an outcome within a sample set as a basis for determining the probability of that outcome.
We take two events, one with conditions and other one without conditions. The later one is called the universal event which chooses all possible options.
We know that a pair of fair dice is thrown 3 times. We take the conditional event A as getting a score of exactly 9 in a throw and the universal event U as throwing the pair of dice and numbers will be denoted as $n\left( A \right)$, $n\left( U \right)$ respectively.
Every dice can show numbers from 1 to 6. So, it has 6 options. For each throw of the pair of the dice the number of outcomes will be $6\times 6=36$. So, $n\left( U \right)=36$
Now a sum of 9 for two dice can be achieved from the combinations of $3\And 6$ and $4\And 5$.
The possible choices for each throw will be $2!+2!=4$. The combinations are $\left( 3,6 \right)$, $\left( 4,5 \right)$, $\left( 5,4 \right)$, $\left( 6,3 \right)$. So, $n\left( A \right)=4$.
We take the empirical probability of getting 9 as sum as $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( U \right)}=\dfrac{4}{36}=\dfrac{1}{9}$. The complement probability of not getting 9 will be $1-\dfrac{1}{9}=\dfrac{8}{9}$.
Now we consider getting 2 out of 3 throws. These events are independent events.
The empirical probability of the event is ${}^{3}{{C}_{2}}\times \dfrac{1}{9}\times \dfrac{1}{9}\times \dfrac{8}{9}=\dfrac{8}{243}$.
And hence the correct answer is option D.

Note:
We need to understand the concept of universal events. This will be the main event that is implemented before the conditional event. The sum being 9 for two cases does not change the outcomes of the remaining event of not getting a sum of 9.