To solve this problem, we'll use the binomial probability formula. The binomial distribution models the number of successes in a fixed number of independent trials, each with the same probability of success.
The formula for the binomial probability [tex]\( P(X = k) \)[/tex] is:
[tex]\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \][/tex]
Where:
- [tex]\( n \)[/tex] is the number of trials (in this case, the number of voters selected, which is 12),
- [tex]\( k \)[/tex] is the number of successes (in this case, the number of voters who would vote by internet, which is 9),
- [tex]\( p \)[/tex] is the probability of success on a single trial (in this case, the probability that a voter would vote by internet, which is 0.39),
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, calculated as [tex]\( \frac{n!}{k!(n-k)!} \)[/tex].
Given the values:
- [tex]\( n = 12 \)[/tex],
- [tex]\( k = 9 \)[/tex],
- [tex]\( p = 0.39 \)[/tex],
we substitute these into the formula to find the probability:
[tex]\[ P(X = 9) = \binom{12}{9} \cdot 0.39^9 \cdot (1-0.39)^{12-9} \][/tex]
Simplifying, we get:
[tex]\[ P(X = 9) = \binom{12}{9} \cdot 0.39^9 \cdot 0.61^3 \][/tex]
To get the binomial coefficient [tex]\( \binom{12}{9} \)[/tex]:
[tex]\[ \binom{12}{9} = \frac{12!}{9!(12-9)!} = \frac{12!}{9! \cdot 3!} \][/tex]
Using this formula and performing the calculations for the binomial coefficient and the probabilities, we find:
[tex]\[ \binom{12}{9} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \][/tex]
So:
[tex]\[ P(X = 9) = 220 \cdot 0.39^9 \cdot 0.61^3 \][/tex]
Finally, calculating these values together gives us:
[tex]\[ P(X = 9) \approx 0.01042 \][/tex]
Therefore, the probability that exactly 9 of the 12 selected voters would vote by internet is [tex]\( \boxed{0.01042} \)[/tex].
This result is rounded to five decimal places, as required.
