To determine the probability of getting exactly 3 fours when a fair die is rolled 6 times, we can use the binomial distribution formula. The binomial distribution is used in scenarios where there are a fixed number of independent trials, each with two possible outcomes (success or failure).
In this case:
- The number of trials ([tex]\(n\)[/tex]) is 6.
- The number of successful outcomes ([tex]\(k\)[/tex]) is 3 (getting exactly 3 fours).
- The probability of success on a single trial ([tex]\(p\)[/tex]) is [tex]\(\frac{1}{6}\)[/tex] (since there is one four out of six possible outcomes on a die).
The binomial probability formula is given by:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \][/tex]
Where:
- [tex]\(\binom{n}{k}\)[/tex] is the binomial coefficient, calculated as [tex]\(\frac{n!}{k!(n-k)!}\)[/tex].
- [tex]\(p^k\)[/tex] is the probability of success raised to the power of the number of successful outcomes.
- [tex]\((1 - p)^{n - k}\)[/tex] is the probability of failure raised to the power of the number of failures.
Step-by-step Solution:
1. Calculate the binomial coefficient [tex]\(\binom{6}{3}\)[/tex]:
[tex]\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \][/tex]
2. Calculate [tex]\(p^k\)[/tex] and [tex]\((1 - p)^{n - k}\)[/tex]:
[tex]\[ p = \frac{1}{6} \][/tex]
[tex]\[ k = 3 \][/tex]
[tex]\[ p^k = \left(\frac{1}{6}\right)^3 = \frac{1}{216} \][/tex]
[tex]\[ 1 - p = \frac{5}{6} \][/tex]
[tex]\[ n - k = 6 - 3 = 3 \][/tex]
[tex]\[ (1 - p)^{n - k} = \left(\frac{5}{6}\right)^3 = \frac{125}{216} \][/tex]
3. Substitute all the values back into the binomial probability formula:
[tex]\[ P(X = 3) = 20 \times \left(\frac{1}{216}\right) \times \left(\frac{125}{216}\right) = 20 \times \frac{125}{46656} = \frac{2500}{46656} \approx 0.05358 \][/tex]
The probability, rounded to the nearest thousandth, is:
[tex]\[ \boxed{0.054} \][/tex]
