Senem is auditing the tax returns of individuals in three zip codes. Thus far, \( 93.1 \% \) of the audited tax returns have passed the audit. If Senem takes 15 tax returns, what is the probability that at most 14 returns will pass the audit? - Round your answer to 3 decimal places.
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To solve the problem, we can use the binomial probability formula. The probability of exactly \( k \) successes (returns passing the audit) in \( n \) trials (total returns audited) can be calculated as: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where: - \( n = 15 \) (the total number of returns), - \( k \) is the number of successes ranging from 0 to 14, - \( p = 0.931 \) (the probability of a return passing the audit), - \( \binom{n}{k} \) is the binomial coefficient. We need to calculate the probability that at most 14 returns will pass the audit, which means we need to find: \[ P(X \leq 14) = P(X = 0) + P(X = 1) + ... + P(X = 14) \] This can be simplified since: \[ P(X \leq 14) = 1 - P(X = 15) \] Now, let's compute \( P(X = 15) \): \[ P(X = 15) = \binom{15}{15} (0.931)^{15} (1 - 0.931)^{0} \] Calculating this: \[ P(X = 15) = 1 \times (0.931)^{15} \times 1 = (0.931)^{15} \approx 0.397 \] Now, we can find \( P(X \leq 14) \): \[ P(X \leq 14) = 1 - P(X = 15) \approx 1 - 0.397 \approx 0.603 \] Thus, rounding to three decimal places, the probability that at most 14 returns will pass the audit is: \[ \boxed{0.603} \]
