United Airlines' flights from Chicago to Atlanta are on time 70% of the time. Suppose ten flights are randomly selected, and the number of on-time flights is recorded. Round all answers to 3 decimal places. 1. The probability that at least two flights are on time is: square 2. The probability that at most six flights are on time is: square 3. The probability that exactly two flights are on time is: square

To solve these problems, we can use the binomial probability formula. The binomial probability formula is given by:<br /><br />\[<br />P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}<br />\]<br /><br />where:<br />- \( n \) is the number of trials (in this case, 10 flights),<br />- \( k \) is the number of successful trials (on-time flights),<br />- \( p \) is the probability of success on a single trial (0.70 for an on-time flight),<br />- \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).<br /><br />Let's calculate each probability:<br /><br />1. **The probability that at least two flights are on time:**<br /><br />This is equivalent to finding \( P(X \geq 2) \). We can find it by calculating the complement of the probability that fewer than two flights are on time:<br /><br />\[<br />P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1))<br />\]<br /><br />Calculate \( P(X = 0) \):<br /><br />\[<br />P(X = 0) = \binom{10}{0} (0.70)^0 (0.30)^{10} = 1 \times 1 \times 0.0000059049 = 0.000006<br />\]<br /><br />Calculate \( P(X = 1) \):<br /><br />\[<br />P(X = 1) = \binom{10}{1} (0.70)^1 (0.30)^9 = 10 \times 0.7 \times 0.00019683 = 0.001377<br />\]<br /><br />Now, calculate \( P(X \geq 2) \):<br /><br />\[<br />P(X \geq 2) = 1 - (0.000006 + 0.001377) = 1 - 0.001383 = 0.998617<br />\]<br /><br />Rounded to three decimal places, the probability is \( \boxed{0.999} \).<br /><br />2. **The probability that at most six flights are on time:**<br /><br />This is equivalent to finding \( P(X \leq 6) \). We need to sum the probabilities from \( X = 0 \) to \( X = 6 \).<br /><br />\[<br />P(X \leq 6) = \sum_{k=0}^{6} P(X = k)<br />\]<br /><br />We already have \( P(X = 0) \) and \( P(X = 1) \). Let's calculate the rest:<br /><br />Calculate \( P(X = 2) \):<br /><br />\[<br />P(X = 2) = \binom{10}{2} (0.70)^2 (0.30)^8 = 45 \times 0.49 \times 0.0006561 = 0.010817<br />\]<br /><br />Calculate \( P(X = 3) \):<br /><br />\[<br />P(X = 3) = \binom{10}{3} (0.70)^3 (0.30)^7 = 120 \times 0.343 \times 0.002187 = 0.036756<br />\]<br /><br />Calculate \( P(X = 4) \):<br /><br />\[<br />P(X = 4) = \binom{10}{4} (0.70)^4 (0.30)^6 = 210 \times 0.2401 \times 0.00729 = 0.102919<br />\]<br /><br />Calculate \( P(X = 5) \):<br /><br />\[<br />P(X = 5) = \binom{10}{5} (0.70)^5 (0.30)^5 = 252 \times 0.16807 \times 0.0243 = 0.200121<br />\]<br /><br />Calculate \( P(X = 6) \):<br /><br />\[<br />P(X = 6) = \binom{10}{6} (0.70)^6 (0.30)^4 = 210 \times 0.117649 \times 0.081 = 0.250202<br />\]<br /><br />Now, sum them up:<br /><br />\[<br />P(X \leq 6) = 0.000006 + 0.001377 + 0.010817 + 0.036756 + 0.102919 + 0.200121 + 0.250202 = 0.602198<br />\]<br /><br />Rounded to three decimal places, the probability is \( \boxed{0.602} \).<br /><br />3. **The probability that exactly two flights are on time:**<br /><br />We already calculated this in part 1:<br /><br />\[<br />P(X = 2) = 0.010817<br />\]<br /><br />Rounded to three decimal places, the probability is \( \boxed{0.011} \).