Question

a new airline flies airplanes with two different seating configurations and has a policy where customers are randomly assigned seats. dan prefers an aisle seat when he flies, so he compares the probabilities of being assigned an aisle seat in each of the two configurations. which statement is true? a. dan is equally likely to get an aisle seat with whichever configuration his flight uses. b. dan should hope his flight uses configuration a. c. there is not enough information to determine which configuration dan should hope his flight uses. d. dan should hope his flight uses configuration b. configuration a configuration b

Explanation:

Step1: Count aisle seats in A

Configuration A: Rows 6 - 25 (20 rows) with 2 aisle seats per row (left and right). So aisle seats: \(20\times2 = 40\). Total seats: Let's see, rows 6 - 25: 20 rows, 4 seats per row (2 columns each side? Wait, the diagram: Configuration A has 2 columns on left, 2 on right? Wait, no, looking at the diagram, Configuration A: rows 6 - 25 (20 rows) with 4 seats per row (2 left, 2 right), and rows 17 - 25? Wait, no, the labels: Configuration A: rows 6 - 25 (20 rows) and 17 - 25? Wait, maybe better: Configuration A: number of rows: from 6 - 25 is 20 rows (25 - 6 + 1 = 20), and each row has 4 seats (2 left, 2 right), but aisle seats are the ones at the ends (left and right aisles). Wait, actually, in airplane seating, aisle seats are next to the aisle. So Configuration A: two aisles? No, Configuration A: the diagram shows two sections? Wait, no, Configuration A: rows 6 - 25 (20 rows) with 4 seats per row (2 left, 2 right), and aisle seats are the outermost (left aisle: 1 per row, right aisle: 1 per row). So 2 aisle seats per row. So 20 rows × 2 = 40 aisle seats. Total seats: Let's count total seats. Configuration A: rows 6 - 25: 20 rows, 4 seats per row: 20×4 = 80. Wait, no, maybe rows 1 - 5? No, the labels start at 6 - 25 and 17 - 25? Wait, maybe I misread. Wait, Configuration A: the left side has rows 6 - 25 (20 rows) and the right side? No, the diagram: Configuration A has a section with rows 6 - 25 (20 rows) and another? Wait, maybe better to look at Configuration B. Configuration B: rows 9 - 32 (24 rows) and 9 - 21? Wait, no, Configuration B: rows 9 - 32 (24 rows) with 6 seats per row (3 left, 3 right), so aisle seats: 2 per row (left and right aisles). So 24 rows × 2 = 48 aisle seats. Total seats: 24 rows × 6 = 144? Wait, no, rows 9 - 32: 32 - 9 + 1 = 24 rows. So Configuration A: total seats: let's see, rows 6 - 25: 20 rows, 4 seats per row: 80. Configuration B: rows 9 - 32: 24 rows, 6 seats per row: 144. Wait, no, maybe I messed up. Wait, the problem is about probability: probability of aisle seat is (number of aisle seats) / (total number of seats).

Wait, let's re-express:

Configuration A:
- Aisle seats: Let's count the number of aisle seats. Looking at the diagram, Configuration A has two aisles? No, Configuration A: the left side has rows 6 - 25 (20 rows) with 2 aisle seats per row (left and right), so 20×2 = 40 aisle seats.
- Total seats: Each row in Configuration A has 4 seats (2 left, 2 right), so 20 rows × 4 = 80 total seats.

Configuration B:
- Aisle seats: Rows 9 - 32 (24 rows) with 2 aisle seats per row (left and right), so 24×2 = 48 aisle seats.
- Total seats: Each row has 6 seats (3 left, 3 right), so 24 rows × 6 = 144 total seats.

Now, probability of aisle seat in A: 40/80 = 0.5.

Probability in B: 48/144 = 0.333...

Wait, that can't be. Wait, maybe I counted total seats wrong. Wait, Configuration A: maybe rows 1 - 25? No, the labels start at 6 - 25 and 17 - 25. Wait, maybe Configuration A has two sections: rows 6 - 16 (11 rows) and 17 - 25 (9 rows). Wait, 11 + 9 = 20 rows. Each row has 4 seats (2 left, 2 right). So total seats: 20×4 = 80. Aisle seats: 2 per row, so 40.

Configuration B: rows 9 - 21 (13 rows) and 9 - 32 (24 rows)? No, Configuration B: rows 9 - 32 (24 rows) with 6 seats per row (3 left, 3 right), so total seats: 24×6 = 144. Aisle seats: 2 per row, so 48.

Now, probability in A: 40/80 = 0.5.

Probability in B: 48/144 = 1/3 ≈ 0.333.

Wait, that would mean Configuration A has a higher probability of aisle seat (0.5 vs 0.333). So Dan prefers aisle seats, so he should hope for Configuration A (higher probability). But wait, maybe my counting is wrong.

Wait, maybe Configuration A has 2 aisle seats per row, total rows: let's count the number of rows in Configuration A. The labels: Configuration A has rows 6 - 25 (20 rows) and 17 - 25? No, the diagram: Configuration A's left side has rows 6 - 25 (20 rows) and the right side? Wait, no, the problem is about two configurations: A and B. Let's look at the number of aisle seats:

In Configuration A: the number of aisle seats: looking at the diagram, each row has 2 aisle seats (left and right). The number of rows: from 6 to 25, that's 20 rows (25 - 6 + 1 = 20). So 20 rows × 2 = 40 aisle seats.

Total seats in A: each row has 4 seats (2 left, 2 right), so 20 × 4 = 80.

In Configuration B: rows from 9 to 32, that's 24 rows (32 - 9 + 1 = 24). Each row has 6 seats (3 left, 3 right), so 24 × 6 = 144 total seats. Aisle seats: 2 per row (left and right), so 24 × 2 = 48.

Now, probability of aisle seat in A: 40/80 = 0.5.

Probability in B: 48/144 = 1/3 ≈ 0.333.

So probability in A is higher (0.5 > 1/3). Therefore, Dan prefers aisle seats, so he should hope for Configuration A (higher probability of aisle seat). So the correct answer is B.

Step2: Verify

Check the probabilities:

- Configuration A: aisle seats = 40, total seats = 80, probability = 40/80 = 0.5.
- Configuration B: aisle seats = 48, total seats = 144, probability = 48/144 = 1/3 ≈ 0.333.

Since 0.5 > 1/3, Dan (who prefers aisle seats) should hope for Configuration A, so option B is correct.

Answer:

B. Dan should hope his flight uses configuration A