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Calculating and Comparing Probabilities

A veterinarian surveys her clients and finds that 32 percent of the households have dogs, 25 percent have cats, and 11 percent have both dogs and cats. Let event [tex]C[/tex] be choosing a client who has cats and let event [tex]D[/tex] be choosing a client who has dogs. Which statements are true? Check all that apply.

- [tex]P(C \mid D)=0.78[/tex]
- [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]
- [tex]P(C \mid D)=P(D \mid C)[/tex]

Sagot :

GinnyAnswer
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]
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