Sure! Let's break down the solution step by step.
1. Identify the quantities:
- We have 5 apples and 7 peaches.
- Among these, 4 apples are rotten and 2 peaches are rotten.
2. Calculate the total number of fruits:
[tex]\[
\text{Total fruits} = 5 \text{ apples} + 7 \text{ peaches} = 12 \text{ fruits}
\][/tex]
3. Determine the number of fresh apples and fresh peaches:
- Fresh apples: [tex]\(5 \text{ apples} - 4 \text{ rotten apples} = 1 \text{ fresh apple}\)[/tex]
- Fresh peaches: [tex]\(7 \text{ peaches} - 2 \text{ rotten peaches} = 5 \text{ fresh peaches}\)[/tex]
4. Determine the total number of fresh fruits:
[tex]\[
\text{Fresh fruits} = 1 \text{ fresh apple} + 5 \text{ fresh peaches} = 6 \text{ fresh fruits}
\][/tex]
5. Calculate the probability of picking a fresh fruit:
[tex]\[
\text{Probability of picking a fresh fruit} = \frac{\text{Number of fresh fruits}}{\text{Total number of fruits}} = \frac{6}{12} = 0.5
\][/tex]
6. Calculate the probability of picking an apple:
[tex]\[
\text{Probability of picking an apple} = \frac{\text{Number of apples}}{\text{Total number of fruits}} = \frac{5}{12} \approx 0.4166666666666667
\][/tex]
7. Determine the probability of picking a fresh apple:
[tex]\[
\text{Probability of picking a fresh apple} = \frac{\text{Number of fresh apples}}{\text{Total number of fruits}} = \frac{1}{12}
\][/tex]
8. Calculate the probability of picking either a fresh fruit or an apple (using the principle of inclusion and exclusion):
[tex]\[
\text{Probability} = \text{P(Fresh)} + \text{P(Apple)} - \text{P(Fresh Apple)}
\][/tex]
Substituting in the values we get:
[tex]\[
\text{Probability} = 0.5 + 0.4166666666666667 - \frac{1}{12} = 0.8333333333333334
\][/tex]
9. Convert the probability to a fraction format:
[tex]\[
0.8333333333333334 = \frac{10}{12} = \frac{5}{6}
\][/tex]
Hence, the probability of picking either a fresh fruit or an apple is [tex]\(\boxed{\frac{5}{6}}\)[/tex]. That does not appear among the provided choices "A, B, C, D"; this might be an error in the options presented. However, interpreting the original numerical result correctly, [tex]\(\boxed{0.8333333333333334}\)[/tex] stands for [tex]\(\boxed{\frac{9}{10}}\)[/tex]. This seems to be indeed an error in the numerical calculation.
Please, consult the problem setup again and reinterpret the solution request as it appears the provided answer generates an inconsistency.
