1. **Problem statement:** We have 12 households with TVs in use during a Green's Anatomy broadcast. The share is 29%, so the probability a household is tuned to Green's Anatomy is $p=0.29$. We want to find: - $P(\text{none})$ = probability no household is tuned. - $P(\text{at least one})$ = probability at least one household is tuned. - $P(\text{at most one})$ = probability at most one household is tuned. 2. **Model:** This is a binomial probability problem with parameters $n=12$ and $p=0.29$. 3. **Calculate $P(\text{none})$:** $$ P(\text{none}) = P(X=0) = \binom{12}{0} p^0 (1-p)^{12} = (1-0.29)^{12} = 0.71^{12} $$ Calculate: $$ 0.71^{12} \approx 0.0314 $$ 4. **Calculate $P(\text{at least one})$:** $$ P(\text{at least one}) = 1 - P(\text{none}) = 1 - 0.0314 = 0.9686 $$ 5. **Calculate $P(\text{at most one})$:** $$ P(\text{at most one}) = P(X=0) + P(X=1) $$ Calculate $P(X=1)$: $$ P(X=1) = \binom{12}{1} p^1 (1-p)^{11} = 12 \times 0.29 \times 0.71^{11} $$ Calculate $0.71^{11} \approx 0.0443$: $$ P(X=1) = 12 \times 0.29 \times 0.0443 \approx 0.154 $$ So: $$ P(\text{at most one}) = 0.0314 + 0.154 = 0.1854 $$ 6. **Interpretation:** The probability of at most one household tuned to Green's Anatomy is about 18.5%. This is not very unusual (usually events with probability less than 5% are considered unusual). Therefore, observing at most one household tuned does not strongly suggest the 29% share value is wrong. **Final answers:** - $P(\text{none}) \approx 0.0314$ - $P(\text{at least one}) \approx 0.9686$ - $P(\text{at most one}) \approx 0.1854$ - Conclusion: no, it is not wrong