Answer :
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]
