Question

it is estimated that 93% of senior citizens suffer from sleep disorders and 71% suffer from anxiety. moreover, 6% of senior citizens suffer from both sleep disorders and anxiety.
(a) given that a senior citizen suffers from anxiety, what is the probability that he or she also suffers from a sleep disorder? round your answer to the nearest hundredth.

(b) find the probability that a senior citizen suffers from anxiety, given that he or she has a sleep disorder. round your answer to the nearest hundredth.

Explanation:

Step1: Recall Conditional Probability Formula

The formula for conditional probability is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( P(A|B) \) is the probability of event \( A \) given event \( B \), \( P(A \cap B) \) is the probability of both \( A \) and \( B \) occurring, and \( P(B) \) is the probability of event \( B \).

Step2: Solve Part (a)

Let \( A \) be the event of having a sleep disorder and \( B \) be the event of having anxiety. We know \( P(A) = 0.93 \), \( P(B) = 0.71 \), and \( P(A \cap B) = 0.6 \). For part (a), we need \( P(A|B) \). Using the formula:
\( P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.6}{0.71} \approx 0.85 \) (rounded to the nearest hundredth).

Step3: Solve Part (b)

For part (b), we need \( P(B|A) \), the probability of having anxiety given a sleep disorder. Using the conditional probability formula:
\( P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.6}{0.93} \approx 0.65 \) (rounded to the nearest hundredth).

Answer:

(a) \(\approx 0.85\)
(b) \(\approx 0.65\)