Based on a survey, \( 37 \% \) of likely voters would be willing to vote by internet instead of the in-person traditional method of voting. For each of the following, assume that 14 likely voters are randomly selected. Complete parts (a) through (c) below. a. What is the probability that exactly 11 of those selected would do internet voting? (Round to five decimal places as needed.)

We are given that the number of likely voters who would vote by internet follows a binomial distribution with parameters \[ n = 14 \quad \text{and} \quad p = 0.37. \] Let \( X \) be the number of individuals who would vote by internet. The probability mass function is \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}. \] For part (a), we need the probability that exactly 11 out of 14 selected voters would vote by internet voting. Thus, we substitute \( k = 11 \) and \( n = 14 \) into the formula: \[ P(X=11) = \binom{14}{11} (0.37)^{11} (0.63)^3. \] Step 1. Compute the combination: \[ \binom{14}{11} = \binom{14}{3} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 364. \] Step 2. Write the probability expression: \[ P(X=11) = 364 \times (0.37)^{11} \times (0.63)^3. \] Step 3. Evaluating the powers (using a calculator or appropriate software for precision): \[ (0.37)^{11} \approx 1.772 \times 10^{-5} \] \[ (0.63)^3 \approx 0.25005 \] Step 4. Combine the factors: \[ P(X=11) \approx 364 \times 1.772 \times 10^{-5} \times 0.25005 \approx 0.001611. \] Rounding to five decimal places, the probability is approximately \[ P(X=11) \approx 0.00161. \]