A new surgery is successful \( 70 \% \) of the time. If the results of 10 such surgeries are randomly sampled, what is the probability that at most 8 of them are successful? carry your intermediate computations to at least four decimal places, and round your answer to two decimal places. (If necescary, consult a list of formulas.)

To find the probability that at most 8 surgeries are successful out of 10, we can use the binomial probability formula, which is defined as: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where: - \( n \) is the number of trials (in this case, 10 surgeries), - \( k \) is the number of successful trials, - \( p \) is the probability of success on an individual trial (0.70). We need to find the probability for \( k = 0, 1, 2, 3, 4, 5, 6, 7, 8 \) and then sum these probabilities. 1. **Finding probabilities for each \( k \)** using \( n = 10 \) and \( p = 0.70 \): - For \( k = 0 \): \[ P(X = 0) = \binom{10}{0} (0.70)^0 (0.30)^{10} = 1 \cdot 1 \cdot 0.0000059 \approx 0.0000059 \] - For \( k = 1 \): \[ P(X = 1) = \binom{10}{1} (0.70)^1 (0.30)^9 = 10 \cdot 0.70 \cdot 0.00019683 \approx 0.001376 \] - For \( k = 2 \): \[ P(X = 2) = \binom{10}{2} (0.70)^2 (0.30)^8 = 45 \cdot 0.49 \cdot 0.0006561 \approx 0.014210 \] - For \( k = 3 \): \[ P(X = 3) = \binom{10}{3} (0.70)^3 (0.30)^7 = 120 \cdot 0.343 \cdot 0.0024 \approx 0.098539 \] - For \( k = 4 \): \[ P(X = 4) = \binom{10}{4} (0.70)^4 (0.30)^6 = 210 \cdot 0.2401 \cdot 0.0081 \approx 0.394912 \] - For \( k = 5 \): \[ P(X = 5) = \binom{10}{5} (0.70)^5 (0.30)^5 = 252 \cdot 0.16807 \cdot 0.024576 \approx 0.259292 \] - For \( k = 6 \): \[ P(X = 6) = \binom{10}{6} (0.70)^6 (0.30)^4 = 210 \cdot 0.117649 \cdot 0.081 \approx 0.195114 \] - For \( k = 7 \): \[ P(X = 7) = \binom{10}{7} (0.70)^7 (0.30)^3 = 120 \cdot 0.0823543 \cdot 0.027 \approx 0.266278 \] - For \( k = 8 \): \[ P(X = 8) = \binom{10}{8} (0.70)^8 (0.30)^2 = 45 \cdot 0.05764801 \cdot 0.09 \approx 0.23364 \] 2. **Summing the probabilities for \( k = 0 \) to \( k = 8 \)**: \[ P(X \leq 8) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) \] Now, summing these calculated values yields: \[ P(X \leq 8) \approx 0.0000059 + 0.001376 + 0.014210 + 0.098539 + 0.394912 + 0.259292 + 0.195114 + 0.266278 + 0.23364 \approx 1.020372 \] Since the sum of probabilities must equal 1 in probability distributions, we find: \[ P(X \leq 8) = 1 - P(X=9) - P(X