Introduction to Probability with Multiple Dice
When we roll multiple dice, each die acts as an independent event. To find the probability of a specific outcome, we look at the total number of possible outcomes versus the number of favorable outcomes. For three dice, the total number of outcomes is $6 \times 6 \times 6 = 216$.
The Problem
We are rolling three standard six-sided dice and need to find:
- The probability of getting exactly 3 sixes.
- The probability of getting exactly 2 sixes.
Step-by-Step Solution
1. Exactly 3 sixes
For all three dice to show a six, there is only one specific sequence: (6, 6, 6).
- Number of favorable outcomes: $1$
- Total outcomes: $216$
- Probability: $P(3 \text{ sixes}) = \frac{1}{216}$
2. Exactly 2 sixes
This is a binomial probability problem. We want 2 successes (rolling a 6) and 1 failure (rolling something other than a 6).
- Probability of success $p = \frac{1}{6}$
- Probability of failure $q = \frac{5}{6}$
- Number of ways to arrange two 6s and one non-6: $\binom{3}{2} = 3$ (The non-6 can be on the first, second, or third die).
The formula is $P(X=k) = \binom{n}{k} p^k q^{n-k}$: $P(X=2) = \binom{3}{2} \times (\frac{1}{6})^2 \times (\frac{5}{6})^1$ $P(X=2) = 3 \times \frac{1}{36} \times \frac{5}{6} = \frac{15}{216} = \frac{5}{72}$
Intuition
Think of the "2 sixes" case: you could get (6, 6, not-6), (6, not-6, 6), or (not-6, 6, 6). Each of these three combinations has a probability of $(\frac{1}{6}) \times (\frac{1}{6}) \times (\frac{5}{6}) = \frac{5}{216}$. Adding them together gives $\frac{15}{216}$.