(CO 3) Fifty-four percent of US teens have heard of a fax machine. You randomly select 12 US teens. Find the probability that the number of these selected teens that have heard of a fax machine is exactly six (first answer listed below). Find the probability that the number is more than 8 (second answer listed below). \( 0.284,0.160 \) \( 0.217,0.120 \) \( 0.284,0.120 \) \( 0.217,0.280 \)

To solve this problem, we can use the binomial probability formula, which is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where: - \( n \) is the number of trials (in this case, 12 teens), - \( k \) is the number of successes (the number of teens who have heard of a fax machine), - \( p \) is the probability of success (54% or 0.54), - \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \). ### Part 1: Probability that exactly 6 teens have heard of a fax machine Here, \( n = 12 \), \( k = 6 \), and \( p = 0.54 \). Calculating \( P(X = 6) \): \[ P(X = 6) = \binom{12}{6} (0.54)^6 (0.46)^{12-6} \] Calculating \( \binom{12}{6} \): \[ \binom{12}{6} = \frac{12!}{6!6!} = 924 \] Now calculating \( (0.54)^6 \) and \( (0.46)^6 \): \[ (0.54)^6 \approx 0.0457 \] \[ (0.46)^6 \approx 0.0150 \] Now substituting these values into the formula: \[ P(X = 6) = 924 \times 0.0457 \times 0.0150 \approx 0.640 \] ### Part 2: Probability that more than 8 teens have heard of a fax machine To find \( P(X > 8) \), we can calculate \( P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) \). Calculating each of these probabilities using the binomial formula: 1. **For \( k = 9 \)**: \[ P(X = 9) = \binom{12}{9} (0.54)^9 (0.46)^3 \] \[ \binom{12}{9} = \binom{12}{3} = 220 \] \[ (0.54)^9 \approx 0.0247 \] \[ (0.46)^3 \approx 0.0973 \] \[ P(X = 9) \approx 220 \times 0.0247 \times 0.0973 \approx 0.052 \] 2. **For \( k = 10 \)**: \[ P(X = 10) = \binom{12}{10} (0.54)^{10} (0.46)^2 \] \[ \binom{12}{10} = \binom{12}{2} = 66 \] \[ (0.54)^{10} \approx 0.0133 \] \[ (0.46)^2 \approx 0.2116 \] \[ P(X = 10) \approx 66 \times 0.0133 \times 0.2116 \approx 0.184 \] 3. **For \( k = 11 \)**: \[ P(X = 11) = \binom{12}{11} (0.54)^{11} (0.46)^1 \] \[ \binom{12}{11} = 12 \] \[ (0.54)^{11} \approx 0.0072 \] \[ (0.46)^1 \approx 0.46 \] \[ P(X = 11) \approx 12 \times 0.0072 \times 0.46 \approx 0.040 \] 4. **For \( k = 12 \)**: \[ P(X = 12) = \binom{12}{12} (0.54)^{12} (0.46)^0 \] \[ \binom{12}{12} = 1 \] \[ (0.54)^{12} \approx 0.003