Answer :
Let's analyze the given information and calculate the necessary probabilities to determine which statements are true.
Given:
- [tex]\( P(C) = 0.25 \)[/tex] (Probability of having cats)
- [tex]\( P(D) = 0.32 \)[/tex] (Probability of having dogs)
- [tex]\( P(C \cap D) = 0.11 \)[/tex] (Probability of having both cats and dogs)
We are to evaluate the following statements:
1. [tex]\( P(C \mid D) = 0.78 \)[/tex]
2. [tex]\( P(D \mid C) = 0.44 \)[/tex]
3. [tex]\( P(C \cap D) = 0.11 \)[/tex]
4. [tex]\( P(C \cap D) = P(D \cap C) \)[/tex]
5. [tex]\( P(C \mid D) = P(D \mid C) \)[/tex]
### Step-by-Step Calculations:
1. Calculate [tex]\( P(C \mid D) \)[/tex]:
[tex]\( P(C \mid D) = \frac{P(C \cap D)}{P(D)} \)[/tex]
Substituting the given values:
[tex]\( P(C \mid D) = \frac{0.11}{0.32} \approx 0.34375 \)[/tex]
This means [tex]\( P(C \mid D) \approx 0.34375 \)[/tex], not 0.78. Hence, the statement [tex]\( P(C \mid D) = 0.78 \)[/tex] is false.
2. Calculate [tex]\( P(D \mid C) \)[/tex]:
[tex]\( P(D \mid C) = \frac{P(C \cap D)}{P(C)} \)[/tex]
Substituting the given values:
[tex]\( P(D \mid C) = \frac{0.11}{0.25} = 0.44 \)[/tex]
This means [tex]\( P(D \mid C) = 0.44 \)[/tex]. Hence, the statement [tex]\( P(D \mid C) = 0.44 \)[/tex] is true.
3. Verify [tex]\( P(C \cap D) = 0.11 \)[/tex]:
[tex]\( P(C \cap D) = 0.11 \)[/tex]
This verifies that the given probability [tex]\( P(C \cap D) = 0.11 \)[/tex] is correct. Hence, the statement [tex]\( P(C \cap D) = 0.11 \)[/tex] is true.
4. Verify [tex]\( P(C \cap D) = P(D \cap C) \)[/tex]:
By the definition of intersection in probability, [tex]\( P(C \cap D) \)[/tex] is the same as [tex]\( P(D \cap C) \)[/tex].
Hence, the statement [tex]\( P(C \cap D) = P(D \cap C) \)[/tex] is true.
5. Verify [tex]\( P(C \mid D) = P(D \mid C) \)[/tex]:
From the calculations:
- [tex]\( P(C \mid D) \approx 0.34375 \)[/tex]
- [tex]\( P(D \mid C) = 0.44 \)[/tex]
This shows [tex]\( P(C \mid D) \neq P(D \mid C) \)[/tex]. Hence, the statement [tex]\( P(C \mid D) = P(D \mid C) \)[/tex] is false.
### Summary of Statements:
- [tex]\( P(C \mid D) = 0.78 \)[/tex] is false.
- [tex]\( P(D \mid C) = 0.44 \)[/tex] is true.
- [tex]\( P(C \cap D) = 0.11 \)[/tex] is true.
- [tex]\( P(C \cap D) = P(D \cap C) \)[/tex] is true.
- [tex]\( P(C \mid D) = P(D \mid C) \)[/tex] is false.
Thus, the true statements are:
- [tex]\( P(D \mid C) = 0.44 \)[/tex]
- [tex]\( P(C \cap D) = 0.11 \)[/tex]
- [tex]\( P(C \cap D) = P(D \cap C) \)[/tex]
Given:
- [tex]\( P(C) = 0.25 \)[/tex] (Probability of having cats)
- [tex]\( P(D) = 0.32 \)[/tex] (Probability of having dogs)
- [tex]\( P(C \cap D) = 0.11 \)[/tex] (Probability of having both cats and dogs)
We are to evaluate the following statements:
1. [tex]\( P(C \mid D) = 0.78 \)[/tex]
2. [tex]\( P(D \mid C) = 0.44 \)[/tex]
3. [tex]\( P(C \cap D) = 0.11 \)[/tex]
4. [tex]\( P(C \cap D) = P(D \cap C) \)[/tex]
5. [tex]\( P(C \mid D) = P(D \mid C) \)[/tex]
### Step-by-Step Calculations:
1. Calculate [tex]\( P(C \mid D) \)[/tex]:
[tex]\( P(C \mid D) = \frac{P(C \cap D)}{P(D)} \)[/tex]
Substituting the given values:
[tex]\( P(C \mid D) = \frac{0.11}{0.32} \approx 0.34375 \)[/tex]
This means [tex]\( P(C \mid D) \approx 0.34375 \)[/tex], not 0.78. Hence, the statement [tex]\( P(C \mid D) = 0.78 \)[/tex] is false.
2. Calculate [tex]\( P(D \mid C) \)[/tex]:
[tex]\( P(D \mid C) = \frac{P(C \cap D)}{P(C)} \)[/tex]
Substituting the given values:
[tex]\( P(D \mid C) = \frac{0.11}{0.25} = 0.44 \)[/tex]
This means [tex]\( P(D \mid C) = 0.44 \)[/tex]. Hence, the statement [tex]\( P(D \mid C) = 0.44 \)[/tex] is true.
3. Verify [tex]\( P(C \cap D) = 0.11 \)[/tex]:
[tex]\( P(C \cap D) = 0.11 \)[/tex]
This verifies that the given probability [tex]\( P(C \cap D) = 0.11 \)[/tex] is correct. Hence, the statement [tex]\( P(C \cap D) = 0.11 \)[/tex] is true.
4. Verify [tex]\( P(C \cap D) = P(D \cap C) \)[/tex]:
By the definition of intersection in probability, [tex]\( P(C \cap D) \)[/tex] is the same as [tex]\( P(D \cap C) \)[/tex].
Hence, the statement [tex]\( P(C \cap D) = P(D \cap C) \)[/tex] is true.
5. Verify [tex]\( P(C \mid D) = P(D \mid C) \)[/tex]:
From the calculations:
- [tex]\( P(C \mid D) \approx 0.34375 \)[/tex]
- [tex]\( P(D \mid C) = 0.44 \)[/tex]
This shows [tex]\( P(C \mid D) \neq P(D \mid C) \)[/tex]. Hence, the statement [tex]\( P(C \mid D) = P(D \mid C) \)[/tex] is false.
### Summary of Statements:
- [tex]\( P(C \mid D) = 0.78 \)[/tex] is false.
- [tex]\( P(D \mid C) = 0.44 \)[/tex] is true.
- [tex]\( P(C \cap D) = 0.11 \)[/tex] is true.
- [tex]\( P(C \cap D) = P(D \cap C) \)[/tex] is true.
- [tex]\( P(C \mid D) = P(D \mid C) \)[/tex] is false.
Thus, the true statements are:
- [tex]\( P(D \mid C) = 0.44 \)[/tex]
- [tex]\( P(C \cap D) = 0.11 \)[/tex]
- [tex]\( P(C \cap D) = P(D \cap C) \)[/tex]
