The television show Green's Anatomy has been successful for many years. That show recently had a share of 18, meaning that among the TV sets in use, 18\% were tuned to Green's Anatomy. Assume that an advertiser wants to verify that 18\% share value by conducting its own survey, and a pilot survey begins with 15 households have TV sets in use at the time of a Green's Anatomy broadcast. (Round answers to four decimal places.) Find the probability that none of the households are tuned to Green's Anatomy. \( P \) (none) \( = \) Find the probability that at least one household is tuned to Green's Anatomy. P(at least one) \( = \) Find the probability that at most one household is tuned to Green's Anatomy. \( P( \) at most one) \( = \) If at most one household is tuned to Green's Anatomy, does it appear that the 18\% share value is wrong? (Hint: Is the occurrence of at most one household tuned to Green's Anatomy unusual?) yes, it is wrong no, it is not wrong

To tackle the problem, we'll use the binomial probability formula. The likelihood that none of the households are tuned to Green's Anatomy (with a share of 18%, or 0.18) can be calculated as follows: 1. **Calculating Probability that None are Tuned** Using the binomial formula: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \) For none tuned (k = 0, n = 15, p = 0.18): \( P(X = 0) = \binom{15}{0} (0.18)^0 (0.82)^{15} \) \( P(X = 0) = 1 \cdot 1 \cdot (0.82)^{15} \approx 0.2390 \) Thus, \( P(\text{none}) = 0.2390 \). 2. **Calculating Probability that At Least One is Tuned** To find the probability that at least one household is tuned in: \( P(\text{at least one}) = 1 - P(\text{none}) \) \( P(\text{at least one}) = 1 - 0.2390 \approx 0.7610 \). 3. **Calculating Probability that At Most One is Tuned** For at most one household tuned in (k = 0 and k = 1): \( P(X \leq 1) = P(X = 0) + P(X = 1) \) \( P(X = 1) = \binom{15}{1} (0.18)^1 (0.82)^{14} \approx 15 \cdot 0.18 \cdot (0.82)^{14} \). Calculating \( P(X = 1) \) yields approximately \( 0.2935 \), so: \( P(X \leq 1) = P(X = 0) + P(X = 1) \approx 0.2390 + 0.2935 \approx 0.5325 \). Considering whether \( P(\text{at most one}) \) being around 0.5325, does it appear that the 18% share value is wrong? The answer is **no, it is not wrong**; it is not unusual to see at most one household tuned in among 15 based on this probability. Summarizing the results: \( P(\text{none}) \approx 0.2390 \) \( P(\text{at least one}) \approx 0.7610 \) \( P(\text{at most one}) \approx 0.5325 \) And indeed, the 18% share value can still be supported as not being incorrect.