Animated Solution for Mathematics - Probability: A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is

When you throw two fair dice, the first die can land on any of $6$ faces, and the second can also land on any of $6$ faces. By the Fundamental Counting Principle, the total number of outcomes is $6\times 6=36$.

Imagine a $6\times 6$ grid on your desk. Every point $(i,j)$ on this grid represents a unique outcome, forming the foundation of our entire calculation.

Identifying Success

We are hunting for a sum of exactly $9$. Let's systematically list the pairs that satisfy this condition: $(3,6)$, $(4,5)$, $(5,4)$, and $(6,3)$.

There are exactly $4$ favorable outcomes. If we define 'success' as rolling a $9$, the probability of success in a single throw is:

Consequently, the probability of failure—rolling anything other than a $9$—is:

This is our binary reality: in every throw, you either succeed with probability $\frac{1}{9}$ or you fail with probability $\frac{8}{9}$.

Why We Use It

We are throwing the dice three times, which constitutes three independent trials. We want success exactly twice.

Because the order of successes and failures matters, we invoke the Binomial Distribution, $X\sim B(n,p)$, where $n=3$ and $r=2$. The Binomial Formula is our bridge:

The term $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ (or '$n$ choose $r$') is the genius of the formula; it calculates exactly how many ways we can arrange those two successes among the three trials.

Bringing It Home

Now, we substitute our values into the formula: $n=3$, $r=2$, $p=\frac{1}{9}$, and $q=\frac{8}{9}$. The expression becomes:

First, calculate the combination: $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$. Next, calculate the powers:

Putting it all together, we have:

Finally, we simplify by dividing both the numerator and the denominator by $3$. We arrive at our elegant final answer: