Question

(1 point) Suppose that \( P(A)=0.49, P(C \mid A)=0.0055 \), and \( P\left(C^{\prime} \mid A^{\prime}\right)=0.01 \). Find \( P(A \mid C) \).

Answer 1

The conditional probability
`\( P(A \mid C) \)` is approximately 0.043.

Step-by-step explanation:

To find
`\( P(A \mid C) \)`, we can use Bayes' Theorem, which states:

`\[ P(A \mid C) = (P(C \mid A) \cdot P(A))/(P(C)) \]`

Here,
`\( P(C) \)` is the probability of event C, and it can be expressed using the law of total probability as:

`\[ P(C) = P(C \mid A) \cdot P(A) + P(C \mid A') \cdot P(A') \]`

Given that
`\( P(A) = 0.49 \), \( P(C \mid A) = 0.0055 \), and \( P(C' \mid A') = 0.01 \),` we can substitute these values into the formulas.

`\[ P(C) = (0.0055 \cdot 0.49) + (0.01 \cdot 0.51) \]`

`\[ P(A \mid C) = (0.0055 \cdot 0.49)/((0.0055 \cdot 0.49) + (0.01 \cdot 0.51)) \]`

Calculating these values yields
`\( P(A \mid C) \approx 0.043 \).` This means that given event C has occurred, the probability of event A is approximately 0.043.