Absolutely, let's solve this problem step-by-step:
Step 1: Identify the parameters of the problem.
- A die is rolled four times, which means we have 4 trials ([tex]\( n = 4 \)[/tex]).
- The probability of rolling a 3 on a six-sided die is [tex]\( \frac{1}{6} \)[/tex].
- We want to find the probability of obtaining exactly 2 threes in these 4 rolls.
Step 2: Define the binomial probability formula.
The binomial probability formula is given by:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
Where:
- [tex]\( n \)[/tex] is the number of trials.
- [tex]\( k \)[/tex] is the number of successes (rolling exactly 2 threes in this case).
- [tex]\( p \)[/tex] is the probability of success on an individual trial.
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, which calculates the number of ways to choose k successes in n trials.
Step 3: Substitute the given values into the formula.
- [tex]\( n = 4 \)[/tex]
- [tex]\( k = 2 \)[/tex]
- [tex]\( p = \frac{1}{6} \)[/tex]
- Probability of not rolling a 3 on one roll [tex]\( q = 1 - p = \frac{5}{6} \)[/tex]
Step 4: Calculate the binomial coefficient [tex]\( \binom{4}{2} \)[/tex].
[tex]\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \][/tex]
Step 5: Compute the probability of exactly two 3s using the binomial formula.
[tex]\[ P(X = 2) = 6 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{4-2} \][/tex]
Simplify the expression:
[tex]\[ P(X = 2) = 6 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2 \][/tex]
[tex]\[ P(X = 2) = 6 \left(\frac{1}{36}\right) \left(\frac{25}{36}\right) \][/tex]
[tex]\[ P(X = 2) = 6 \times \frac{25}{1296} \][/tex]
[tex]\[ P(X = 2) = \frac{150}{1296} \][/tex]
[tex]\[ P(X = 2) = 0.1157 \][/tex]
Conclusion
So, the probability of rolling exactly two 3s when a fair die is rolled four times is:
[tex]\[ \boxed{0.1157} \][/tex]
This rounded to four decimal places gives us the desired probability.
