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) $$ =.0510 $$ Find the probability that at least one household is tuned to Green's Anatomy. P(at least one) $$ =1 $$ 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?) O yes, it is wrong Ono, it is not wrong

Question

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Let $$ X $$ be the number of households (out of 15) tuned to Green's Anatomy. We assume
$$
X \sim \text{Binomial}(n=15, p=0.18).
$$

**Step 1.** Find the probability that none of the households are tuned in, i.e. $$ P(X=0) $$.

We have
$$
P(X=0) = \binom{15}{0}(0.18)^0(0.82)^{15} = (0.82)^{15}.
$$
Using a calculator,
$$
(0.82)^{15} \approx 0.0510.
$$

**Step 2.** Find the probability that at least one household is tuned in, i.e. $$ P(X\ge1) $$.

This is the complement of the probability that none are tuned:
$$
P(X\ge1) = 1 - P(X=0) = 1 - (0.82)^{15} \approx 1 - 0.0510 = 0.9490.
$$

**Step 3.** Find the probability that at most one household is tuned in, i.e. $$ P(X\le 1) $$.

This probability is the sum of the probabilities of 0 or 1 household being tuned in:
$$
P(X\le1) = P(X=0) + P(X=1).
$$
We already have $$ P(X=0)= 0.0510 $$. Now, calculate $$ P(X=1) $$:
$$
P(X=1) = \binom{15}{1}(0.18)^1(0.82)^{14} = 15 \times 0.18 \times (0.82)^{14}.
$$
Calculating step by step:
- $$ 15 \times 0.18 = 2.7 $$.
- $$ (0.82)^{14} $$ can be computed as follows. (Using logarithms or a calculator, one finds $$ (0.82)^{14} \approx 0.0620 $$ approximately.)

Thus,
$$
P(X=1) \approx 2.7 \times 0.0620 \approx 0.1674.
$$
Then,
$$
P(X\le1) \approx 0.0510 + 0.1674 = 0.2184.
$$

**Step 4.** Assess if at most one household tuned in is unusual given an $$18\%$$ share.

The probability $$ P(X\le1) \approx 0.2184 $$ means that there is about a $$21.84\%$$ chance of observing at most one household tuned to Green's Anatomy. This value is not extremely low (it is much higher than typical significance levels such as $$5\%$$ or $$1\%$$). In the context of verifying the $$18\%$$ share with just 15 households, observing at most one tuned household is not considered an unusual event.

Thus, if at most one household is tuned to Green's Anatomy, it does not provide strong evidence that the $$18\%$$ share value is wrong.

**Final Answers:**

- $$ P(\text{none}) \approx 0.0510 $$
- $$ P(\text{at least one}) \approx 0.9490 $$
- $$ P(\text{at most one}) \approx 0.2184 $$
- Conclusion: No, it is not wrong.
- The probability that none of the households are tuned to Green's Anatomy is approximately 0.0510. - The probability that at least one household is tuned to Green's Anatomy is approximately 0.9490. - The probability that at most one household is tuned to Green's Anatomy is approximately 0.2184. - Observing at most one household tuned to Green's Anatomy does not indicate that the 18% share value is wrong.

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