If a store is chosen at random at Mayfair Mall, 9. what is the probability of choosing a store that does not sell food or one that is located on the second floor? \( 59 / 61 \) \( 118 / 122 \) \( 61 / 59 \) \( 122 / 118 \)

To find the probability of choosing a store that does not sell food or one that is located on the second floor, we can use the principle of inclusion-exclusion. Let: - \( A \) be the event of choosing a store that does not sell food. - \( B \) be the event of choosing a store that is located on the second floor. We want to find \( P(A \cup B) \), which can be calculated using the formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] However, we need the number of stores in each category to calculate this probability. Since the problem does not provide specific numbers of stores, we can only analyze the options given. Assuming: - The total number of stores is \( N \). - The number of stores that do not sell food is \( n_A \). - The number of stores on the second floor is \( n_B \). - The number of stores that do not sell food and are on the second floor is \( n_{A \cap B} \). Then, the probability can be expressed as: \[ P(A \cup B) = \frac{n_A + n_B - n_{A \cap B}}{N} \] Given the options: 1. \( \frac{59}{61} \) 2. \( \frac{118}{122} \) 3. \( \frac{61}{59} \) 4. \( \frac{122}{118} \) We can simplify the fractions to see if they represent valid probabilities (which must be between 0 and 1): 1. \( \frac{59}{61} \approx 0.967 \) 2. \( \frac{118}{122} \approx 0.967 \) 3. \( \frac{61}{59} \approx 1.034 \) (not valid) 4. \( \frac{122}{118} \approx 1.034 \) (not valid) The valid probabilities are \( \frac{59}{61} \) and \( \frac{118}{122} \), both of which are approximately 0.967. Since we do not have the specific counts of stores, we cannot definitively choose between the two valid options without additional information. However, both \( \frac{59}{61} \) and \( \frac{118}{122} \) are valid probabilities for the event described. If you have more context or specific numbers of stores, please provide that for a more accurate calculation.