Understanding Binomial Probability
The problem asks for the probability of getting exactly two 'successes' in four rolls of a die. This is a classic application of the Binomial Distribution.
The Binomial Formula
The formula to find the probability of exactly $k$ successes in $n$ independent trials is:
$$P(X = k) = \binom{n}{k} p^k q^{(n-k)}$$
Where:
- $n$ = total number of trials (in this case, 4 rolls).
- $k$ = number of successful outcomes (in this case, 2).
- $p$ = probability of success in a single trial.
- $q$ = probability of failure in a single trial ($1 - p$).
- $\binom{n}{k}$ = the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$.
Step-by-Step Solution
1. Define the variables:
- $n = 4$ (rolls)
- $k = 2$ (successes)
- $p = \frac{3}{6} = 0.5$ (since a standard die has even numbers 2, 4, and 6)
- $q = 1 - 0.5 = 0.5$
2. Calculate the binomial coefficient: $$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1) \times (2 \times 1)} = 6$$
3. Plug into the formula: $$P(X = 2) = 6 \times (0.5)^2 \times (0.5)^{(4-2)}$$ $$P(X = 2) = 6 \times (0.5)^2 \times (0.5)^2$$ $$P(X = 2) = 6 \times 0.25 \times 0.25$$ $$P(X = 2) = 6 \times 0.0625 = 0.375$$
Thus, the probability of getting exactly two successes is $0.375$ or $3/8$.