Consider the following scenario: Let $$ \mathrm{P}(C)=0.4 $$. Let $$ \mathrm{P}(D)=0.7 $$. Let $$ \mathrm{P}(C \mid D)=0.32 $$. Note: Give your answers in their exact decimal forms. (a) FindP $$ (C $$ AND $$ D) $$. $$ \mathrm{P}(C $$ AND $$ D)=0.32 $$

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It appears there's a small misunderstanding in calculating $$ \mathrm{P}(C \text{ AND } D) $$. Let's walk through the correct computation step by step.

### Given:
- $$ \mathrm{P}(C) = 0.4 $$
- $$ \mathrm{P}(D) = 0.7 $$
- $$ \mathrm{P}(C \mid D) = 0.32 $$

### To Find:
- $$ \mathrm{P}(C \text{ AND } D) $$

### Understanding the Relationship:
The probability of both events $$ C $$ and $$ D $$ occurring together, denoted as $$ \mathrm{P}(C \text{ AND } D) $$, can be found using the definition of conditional probability:

$$
\mathrm{P}(C \text{ AND } D) = \mathrm{P}(C \mid D) \times \mathrm{P}(D)
$$

### Calculation:
Plugging in the given values:

$$
\mathrm{P}(C \text{ AND } D) = 0.32 \times 0.7 = 0.224
$$

### Conclusion:
$$
\mathrm{P}(C \text{ AND } D) = 0.224
$$

So, the correct probability of both $$ C $$ and $$ D $$ occurring together is **0.224**.
$$ \mathrm{P}(C \text{ AND } D) = 0.224 $$

Consider the following scenario: Let \( \mathrm{P}(C)=0.4 \). Let \( \mathrm{P}(D)=0.7 \). Let \( \mathrm{P}(C \mid D)=0.32 \). Note: Give your answers in their exact decimal forms. (a) FindP \( (C \) AND \( D) \). \( \mathrm{P}(C \) AND \( D)=0.32 \)

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