Questions: Conditional probability: Basic It is estimated that 89% of senior citizens suffer from sleep disorders and 9% suffer from anxiety. Moreover, 5% of senior citizens suffer from both sleep disorders and anxiety. Answer the questions below. (If necessary, consult a list of formulas.) (a) Find the probability that a senior citizen suffers from anxiety, given that he or she has sleep disorder. Round your answer to 2 decimal places. (b) Given that a senior citizen suffers from anxiety, what is the probability that he or she also suffers from sleep disorder? Round your answer to 2 decimal places.

Conditional probability: Basic

It is estimated that 89% of senior citizens suffer from sleep disorders and 9% suffer from anxiety. Moreover, 5% of senior citizens suffer from both sleep disorders and anxiety.
Answer the questions below. (If necessary, consult a list of formulas.)
(a) Find the probability that a senior citizen suffers from anxiety, given that he or she has sleep disorder. Round your answer to 2 decimal places.

(b) Given that a senior citizen suffers from anxiety, what is the probability that he or she also suffers from sleep disorder? Round your answer to 2 decimal places.

Transcript text: Conditional probability: Basic It is estimated that $89 \%$ of senior citizens suffer from sleep disorders and $9 \%$ suffer from anxiety. Moreover, $5 \%$ of senior citizens suffer from both sleep disorders and anxiety. Answer the questions below. (If necessary, consult a list of formulas.) (a) Find the probability that a senior citizen suffers from anxiety, given that he or she has sleep disorder. Round your answer to 2 decimal places. $\square$ (b) Given that a senior citizen suffers from anxiety, what is the probability that he or she also suffers from sleep disorder? Round your answer to 2 decimal places. $\square$

Solution

Solution Steps

Step 1: Identify the given probabilities

Probability of sleep disorder ($ P(A) $) = $ 89\% = 0.89 $.
Probability of anxiety ($ P(B) $) = $ 9\% = 0.09 $.
Probability of both sleep disorder and anxiety ($ P(A \cap B) $) = $ 5\% = 0.05 $.

Step 2: Solve part (a) - Find $ P(B \mid A) $

The conditional probability formula is: \[ P(B \mid A) = \frac{P(A \cap B)}{P(A)} \] Substitute the given values: \[ P(B \mid A) = \frac{0.05}{0.89} \approx 0.0562 \] Round to 2 decimal places: \[ P(B \mid A) \approx 0.06 \]

Step 3: Solve part (b) - Find $ P(A \mid B) $

The conditional probability formula is: \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \] Substitute the given values: \[ P(A \mid B) = \frac{0.05}{0.09} \approx 0.5556 \] Round to 2 decimal places: \[ P(A \mid B) \approx 0.56 \]

Final Answer

(a) $ \boxed{0.06} $
(b) $ \boxed{0.56} $

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