Dive into the world of knowledge and get your queries resolved at IDNLearn.com. Get accurate answers to your questions from our community of experts who are always ready to provide timely and relevant solutions.
Sagot :
To understand the various probabilities involving fans of professional football and car racing, let's walk through each of the given events and their probabilities using the provided information.
1. Calculating [tex]\( P(F \mid C) \)[/tex]:
- Given:
- [tex]\(P(F) = 0.48\)[/tex]: Probability of being a fan of professional football.
- [tex]\(P(C) = 0.12\)[/tex]: Probability of being a fan of car racing.
- [tex]\(P(C \cap F) = 0.09\)[/tex]: Probability of being a fan of both professional football and car racing.
- We need to calculate the conditional probability [tex]\( P(F \mid C) \)[/tex], which is the probability that a person is a fan of professional football given that they are a fan of car racing.
- Formula for conditional probability:
[tex]\[ P(F \mid C) = \frac{P(C \cap F)}{P(C)} \][/tex]
- Substituting the given values:
[tex]\[ P(F \mid C) = \frac{0.09}{0.12} = 0.75 \][/tex]
- So, [tex]\( P(F \mid C) = 0.75 \)[/tex] is true.
2. Calculating [tex]\( P(C \mid F) \)[/tex]:
- We need to calculate the conditional probability [tex]\( P(C \mid F) \)[/tex], which is the probability that a person is a fan of car racing given that they are a fan of professional football.
- Formula for conditional probability:
[tex]\[ P(C \mid F) = \frac{P(C \cap F)}{P(F)} \][/tex]
- Substituting the given values:
[tex]\[ P(C \mid F) = \frac{0.09}{0.48} = 0.1875 \][/tex]
- So, [tex]\( P(C \mid F) = 0.25 \)[/tex] is false. The correct value is [tex]\( 0.1875 \)[/tex].
3. Verifying [tex]\( P(C \cap F) \)[/tex]:
- [tex]\( P(C \cap F) = 0.09 \)[/tex] is given directly as the probability of being a fan of both professional football and car racing.
- So, [tex]\( P(C \cap F) = 0.09 \)[/tex] is true.
4. Checking if [tex]\( P(C \cap F) = P(F \cap C) \)[/tex]:
- By the commutative property of probabilities involving intersections:
[tex]\[ P(C \cap F) = P(F \cap C) \][/tex]
- This is always true for any events [tex]\( C \)[/tex] and [tex]\( F \)[/tex].
- So, [tex]\( P(C \cap F) = P(F \cap C) \)[/tex] is true.
5. Comparing [tex]\( P(C \mid F) = P(F \mid C) \)[/tex]:
- From our calculated values:
[tex]\[ P(C \mid F) = 0.1875 \quad \text{and} \quad P(F \mid C) = 0.75 \][/tex]
- Clearly, [tex]\( P(C \mid F) \neq P(F \mid C) \)[/tex].
- So, [tex]\( P(C \mid F) = P(F \mid C) \)[/tex] is false.
In summary, the three true statements are:
1. [tex]\( P(F \mid C) = 0.75 \)[/tex]
3. [tex]\( P(C \cap F) = 0.09 \)[/tex]
4. [tex]\( P(C \cap F) = P(F \cap C) \)[/tex]
1. Calculating [tex]\( P(F \mid C) \)[/tex]:
- Given:
- [tex]\(P(F) = 0.48\)[/tex]: Probability of being a fan of professional football.
- [tex]\(P(C) = 0.12\)[/tex]: Probability of being a fan of car racing.
- [tex]\(P(C \cap F) = 0.09\)[/tex]: Probability of being a fan of both professional football and car racing.
- We need to calculate the conditional probability [tex]\( P(F \mid C) \)[/tex], which is the probability that a person is a fan of professional football given that they are a fan of car racing.
- Formula for conditional probability:
[tex]\[ P(F \mid C) = \frac{P(C \cap F)}{P(C)} \][/tex]
- Substituting the given values:
[tex]\[ P(F \mid C) = \frac{0.09}{0.12} = 0.75 \][/tex]
- So, [tex]\( P(F \mid C) = 0.75 \)[/tex] is true.
2. Calculating [tex]\( P(C \mid F) \)[/tex]:
- We need to calculate the conditional probability [tex]\( P(C \mid F) \)[/tex], which is the probability that a person is a fan of car racing given that they are a fan of professional football.
- Formula for conditional probability:
[tex]\[ P(C \mid F) = \frac{P(C \cap F)}{P(F)} \][/tex]
- Substituting the given values:
[tex]\[ P(C \mid F) = \frac{0.09}{0.48} = 0.1875 \][/tex]
- So, [tex]\( P(C \mid F) = 0.25 \)[/tex] is false. The correct value is [tex]\( 0.1875 \)[/tex].
3. Verifying [tex]\( P(C \cap F) \)[/tex]:
- [tex]\( P(C \cap F) = 0.09 \)[/tex] is given directly as the probability of being a fan of both professional football and car racing.
- So, [tex]\( P(C \cap F) = 0.09 \)[/tex] is true.
4. Checking if [tex]\( P(C \cap F) = P(F \cap C) \)[/tex]:
- By the commutative property of probabilities involving intersections:
[tex]\[ P(C \cap F) = P(F \cap C) \][/tex]
- This is always true for any events [tex]\( C \)[/tex] and [tex]\( F \)[/tex].
- So, [tex]\( P(C \cap F) = P(F \cap C) \)[/tex] is true.
5. Comparing [tex]\( P(C \mid F) = P(F \mid C) \)[/tex]:
- From our calculated values:
[tex]\[ P(C \mid F) = 0.1875 \quad \text{and} \quad P(F \mid C) = 0.75 \][/tex]
- Clearly, [tex]\( P(C \mid F) \neq P(F \mid C) \)[/tex].
- So, [tex]\( P(C \mid F) = P(F \mid C) \)[/tex] is false.
In summary, the three true statements are:
1. [tex]\( P(F \mid C) = 0.75 \)[/tex]
3. [tex]\( P(C \cap F) = 0.09 \)[/tex]
4. [tex]\( P(C \cap F) = P(F \cap C) \)[/tex]
Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. IDNLearn.com provides the answers you need. Thank you for visiting, and see you next time for more valuable insights.
