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To determine whether being an artist is independent of watching contests for the social media users, we need to compare the joint probability of being an artist and watching contests with the product of the individual probabilities of being an artist and watching contests.
### Step-by-Step Solution
1. Total number of users:
- Given: Total users = 290
2. Number of artists:
- Given: Artists = 31
3. Number of users watching contests:
- Given: Contests viewers = 5 (artists who watch contests)
4. Number of artists watching contests:
- Given: Contest-watching artists = 5
5. Calculate the probability of being an artist:
- [tex]\( P(\text{Artist}) = \frac{\text{Number of artists}}{\text{Total number of users}} \)[/tex]
- [tex]\( P(\text{Artist}) = \frac{31}{290} \)[/tex]
- [tex]\( P(\text{Artist}) \approx 0.1069 \)[/tex]
6. Calculate the probability of watching contests:
- [tex]\( P(\text{Watching Contests}) = \frac{\text{Number of users watching contests}}{\text{Total number of users}} \)[/tex]
- [tex]\( P(\text{Watching Contests}) = \frac{5}{290} \)[/tex]
- [tex]\( P(\text{Watching Contests}) \approx 0.0172 \)[/tex]
7. Calculate the joint probability of being an artist and watching contests:
- [tex]\( P(\text{Artist and Watching Contests}) = \frac{\text{Number of artists watching contests}}{\text{Total number of users}} \)[/tex]
- [tex]\( P(\text{Artist and Watching Contests}) = \frac{5}{290} \)[/tex]
- [tex]\( P(\text{Artist and Watching Contests}) \approx 0.0172 \)[/tex]
8. Calculate the product of the individual probabilities:
- [tex]\( P(\text{Artist}) \times P(\text{Watching Contests}) \)[/tex]
- [tex]\( (0.1069) \times (0.0172) \)[/tex]
- [tex]\( \approx 0.0018 \)[/tex]
9. Compare the joint probability with the product of individual probabilities:
- Joint probability: [tex]\( \approx 0.0172 \)[/tex]
- Product of individual probabilities: [tex]\( \approx 0.0018 \)[/tex]
Since 0.0172 is not equal to 0.0018, the probabilities are not equal. Therefore, being an artist and watching contests are not independent events.
### Conclusion
No, because [tex]\(0.017 \neq 0.0018\)[/tex].
The correct answer is:
No, because [tex]\(0.017 \neq (0.217)(0.207)\)[/tex].
### Step-by-Step Solution
1. Total number of users:
- Given: Total users = 290
2. Number of artists:
- Given: Artists = 31
3. Number of users watching contests:
- Given: Contests viewers = 5 (artists who watch contests)
4. Number of artists watching contests:
- Given: Contest-watching artists = 5
5. Calculate the probability of being an artist:
- [tex]\( P(\text{Artist}) = \frac{\text{Number of artists}}{\text{Total number of users}} \)[/tex]
- [tex]\( P(\text{Artist}) = \frac{31}{290} \)[/tex]
- [tex]\( P(\text{Artist}) \approx 0.1069 \)[/tex]
6. Calculate the probability of watching contests:
- [tex]\( P(\text{Watching Contests}) = \frac{\text{Number of users watching contests}}{\text{Total number of users}} \)[/tex]
- [tex]\( P(\text{Watching Contests}) = \frac{5}{290} \)[/tex]
- [tex]\( P(\text{Watching Contests}) \approx 0.0172 \)[/tex]
7. Calculate the joint probability of being an artist and watching contests:
- [tex]\( P(\text{Artist and Watching Contests}) = \frac{\text{Number of artists watching contests}}{\text{Total number of users}} \)[/tex]
- [tex]\( P(\text{Artist and Watching Contests}) = \frac{5}{290} \)[/tex]
- [tex]\( P(\text{Artist and Watching Contests}) \approx 0.0172 \)[/tex]
8. Calculate the product of the individual probabilities:
- [tex]\( P(\text{Artist}) \times P(\text{Watching Contests}) \)[/tex]
- [tex]\( (0.1069) \times (0.0172) \)[/tex]
- [tex]\( \approx 0.0018 \)[/tex]
9. Compare the joint probability with the product of individual probabilities:
- Joint probability: [tex]\( \approx 0.0172 \)[/tex]
- Product of individual probabilities: [tex]\( \approx 0.0018 \)[/tex]
Since 0.0172 is not equal to 0.0018, the probabilities are not equal. Therefore, being an artist and watching contests are not independent events.
### Conclusion
No, because [tex]\(0.017 \neq 0.0018\)[/tex].
The correct answer is:
No, because [tex]\(0.017 \neq (0.217)(0.207)\)[/tex].
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