From experience, an airline knows that only \( 85 \% \) of the passengers booked for a certain flight actually show up. If 9 passengers are randomly selected, find the probability that at most 6 of them show up. Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places. (If necessary, consult a list of formulas.)

Simplify the expression by following steps: - step0: Evaluate the power: \(0.15^{9}\) - step1: Convert the expressions: \(\left(\frac{3}{20}\right)^{9}\) - step2: Evaluate the power: \(\frac{19683}{20^{9}}\) Expand the expression \( 36 * (0.85)^2 * (0.15)^7 \) Simplify the expression by following steps: - step0: Simplify: \(36\times 0.85^{2}\times 0.15^{7}\) - step1: Convert the expressions: \(36\left(\frac{17}{20}\right)^{2}\times 0.15^{7}\) - step2: Convert the expressions: \(36\left(\frac{17}{20}\right)^{2}\left(\frac{3}{20}\right)^{7}\) - step3: Multiply the terms: \(\frac{2601}{100}\left(\frac{3}{20}\right)^{7}\) - step4: Evaluate the power: \(\frac{2601}{100}\times \frac{3^{7}}{20^{7}}\) - step5: Multiply the fractions: \(\frac{2601\times 3^{7}}{100\times 20^{7}}\) - step6: Multiply: \(\frac{5688387}{100\times 20^{7}}\) Expand the expression \( 84 * (0.85)^6 * (0.15)^3 \) Simplify the expression by following steps: - step0: Simplify: \(84\times 0.85^{6}\times 0.15^{3}\) - step1: Convert the expressions: \(84\left(\frac{17}{20}\right)^{6}\times 0.15^{3}\) - step2: Convert the expressions: \(84\left(\frac{17}{20}\right)^{6}\left(\frac{3}{20}\right)^{3}\) - step3: Simplify: \(84\times \frac{17^{6}}{20^{6}}\times \left(\frac{3}{20}\right)^{3}\) - step4: Multiply the terms: \(\frac{21\times 17^{6}}{16000000}\left(\frac{3}{20}\right)^{3}\) - step5: Evaluate the power: \(\frac{21\times 17^{6}}{16000000}\times \frac{3^{3}}{20^{3}}\) - step6: Multiply the fractions: \(\frac{21\times 17^{6}\times 3^{3}}{16000000\times 20^{3}}\) - step7: Multiply: \(\frac{567\times 17^{6}}{128000000000}\) Expand the expression \( 126 * (0.85)^4 * (0.15)^5 \) Simplify the expression by following steps: - step0: Simplify: \(126\times 0.85^{4}\times 0.15^{5}\) - step1: Convert the expressions: \(126\left(\frac{17}{20}\right)^{4}\times 0.15^{5}\) - step2: Convert the expressions: \(126\left(\frac{17}{20}\right)^{4}\left(\frac{3}{20}\right)^{5}\) - step3: Simplify: \(126\times \frac{17^{4}}{20^{4}}\times \left(\frac{3}{20}\right)^{5}\) - step4: Multiply the terms: \(\frac{63\times 17^{4}}{80000}\left(\frac{3}{20}\right)^{5}\) - step5: Evaluate the power: \(\frac{63\times 17^{4}}{80000}\times \frac{3^{5}}{20^{5}}\) - step6: Multiply the fractions: \(\frac{63\times 17^{4}\times 3^{5}}{80000\times 20^{5}}\) - step7: Multiply: \(\frac{15309\times 17^{4}}{80000\times 20^{5}}\) Expand the expression \( 9 * (0.85)^1 * (0.15)^8 \) Simplify the expression by following steps: - step0: Simplify: \(9\times 0.85^{1}\times 0.15^{8}\) - step1: Calculate: \(9\times \frac{17}{20}\times 0.15^{8}\) - step2: Convert the expressions: \(9\times \frac{17}{20}\left(\frac{3}{20}\right)^{8}\) - step3: Multiply the terms: \(\frac{153}{20}\left(\frac{3}{20}\right)^{8}\) - step4: Evaluate the power: \(\frac{153}{20}\times \frac{3^{8}}{20^{8}}\) - step5: Multiply the fractions: \(\frac{153\times 3^{8}}{20\times 20^{8}}\) - step6: Multiply: \(\frac{1003833}{20^{9}}\) Expand the expression \( 84 * (0.85)^3 * (0.15)^6 \) Simplify the expression by following steps: - step0: Simplify: \(84\times 0.85^{3}\times 0.15^{6}\) - step1: Convert the expressions: \(84\left(\frac{17}{20}\right)^{3}\times 0.15^{6}\) - step2: Convert the expressions: \(84\left(\frac{17}{20}\right)^{3}\left(\frac{3}{20}\right)^{6}\) - step3: Multiply the terms: \(\frac{103173}{2000}\left(\frac{3}{20}\right)^{6}\) - step4: Evaluate the power: \(\frac{103173}{2000}\times \frac{3^{6}}{20^{6}}\) - step5: Multiply the fractions: \(\frac{103173\times 3^{6}}{2000\times 20^{6}}\) - step6: Multiply: \(\frac{75213117}{2000\times 20^{6}}\) Expand the expression \( 126 * (0.85)^5 * (0.15)^4 \) Simplify the expression by following steps: - step0: Simplify: \(126\times 0.85^{5}\times 0.15^{4}\) - step1: Convert the expressions: \(126\left(\frac{17}{20}\right)^{5}\times 0.15^{4}\) - step2: Convert the expressions: \(126\left(\frac{17}{20}\right)^{5}\left(\frac{3}{20}\right)^{4}\) - step3: Simplify: \(126\times \frac{17^{5}}{20^{5}}\times \left(\frac{3}{20}\right)^{4}\) - step4: Multiply the terms: \(\frac{63\times 17^{5}}{1600000}\left(\frac{3}{20}\right)^{4}\) - step5: Evaluate the power: \(\frac{63\times 17^{5}}{1600000}\times \frac{3^{4}}{20^{4}}\) - step6: Multiply the fractions: \(\frac{63\times 17^{5}\times 3^{4}}{1600000\times 20^{4}}\) - step7: Multiply: \(\frac{5103\times 17^{5}}{1600000\times 20^{4}}\) Let's summarize the calculations for each probability \( P(X = k) \) where \( k = 0, 1, 2, 3, 4, 5, 6 \): 1. **For \( k = 0 \)**: \[ P(X = 0) \approx 3.844336 \times 10^{-8} \] 2. **For \( k = 1 \)**: \[ P(X = 1) \approx 1.960611 \times 10^{-6} \] 3. **For \( k = 2 \)**: \[ P(X = 2) \approx 4.444052 \times 10^{-5} \] 4. **For \( k = 3 \)**: \[ P(X = 3) \approx 0.000588 \] 5. **For \( k = 4 \)**: \[ P(X = 4) \approx 0.004995 \] 6. **For \( k = 5 \)**: \[ P(X = 5) \approx 0.028303 \] 7. **For \( k = 6 \)**: \[ P(X = 6) \approx 0.106922 \] ### Total Probability \( P(X \leq 6) \): Now, we will sum these probabilities to find \( P(X \leq 6) \): \[ P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) \] Calculating the total: \[ P(X \leq 6) \approx 3.844336 \times 10^{-8} + 1.960611 \times 10^{-6} + 4.444052 \times 10^{-5} + 0.000588 + 0.004995 + 0.028303 + 0.106922 \] Let's compute this sum. Simplify the expression by following steps: - step0: Simplify: \(3.844336e-8+1.960611e-6+4.444052e-5+0.000588+0.004995+0.028303+0.106922\) - step1: Add the numbers: \(10.248999e-18.859192\) - step2: Simplify: \(27.859668-18.859192\) - step3: Calculate: \(9.000476\) The total probability that at most 6 passengers show up is approximately: \[ P(X \leq 6) \approx 0.9000 \] ### Final Answer: Rounding to two decimal places, the probability that at most 6 of the 9 passengers show up is: \[ \boxed{0.90} \]