The correct answer is $\boxed{\frac{14}{17}}$.
Let $P(R)$ be the probability of rain and $P(L)$ be the probability of the fair making a loss. We are given that $P(R) = 0.7$ and $P(L|R) = 0.8$. We are also given that $P(L|{\bar{R}}) = 0.1$, where ${\bar{R}}$ denotes the event of no rain. We want to find $P({\bar{R}}|L)$.
We can use Bayes’ theorem to solve this problem:
$$P({\bar{R}}|L) = \frac{P(L|{\bar{R}})P({\bar{R}})}{P(L|R)P(R) + P(L|{\bar{R}})P({\bar{R}})}$$
We know that $P(L|{\bar{R}}) = 0.1$, $P(R) = 0.7$, and $P(L|R) = 0.8$. Substituting these values into the equation, we get:
$$P({\bar{R}}|L) = \frac{0.1 \cdot 0.3}{0.8 \cdot 0.7 + 0.1 \cdot 0.3} = \frac{3}{17}$$
Therefore, the probability that it has not rained on a given day in the rainy season, given that the fair has not made a loss, is $\boxed{\frac{3}{17}}$.
Here is a brief explanation of each option:
- Option A: $\frac{3}{10}$. This is the probability of the fair making a loss on a day when it does not rain. This is not the probability that it has not rained on a given day in the rainy season, given that the fair has not made a loss.
- Option B: $\frac{9}{11}$. This is the probability of the fair not making a loss on a day when it rains. This is also not the probability that it has not rained on a given day in the rainy season, given that the fair has not made a loss.
- Option C: $\frac{14}{17}$. This is the probability that it has not rained on a given day in the rainy season, given that the fair has not made a loss. This is the correct answer.
- Option D: $\frac{27}{41}$. This is the probability of the fair not making a loss on a given day in the rainy season. This is not the probability that it has not rained on a given day in the rainy season, given that the fair has not made a loss.