Question

what is the probability that a passenger prefers to sit in the front of the plane and prefers a window seat?
a. \\(\frac{2}{9}\\)
\\(\frac{1}{3}\\)
c. \\(\frac{4}{11}\\)
d. \\(\frac{2}{3}\\)

	front	back	total
window	8	14	22
total	12	24	36

Explanation:

Step1: Identify relevant counts

From the table, the number of passengers who prefer front and window seat is 8. The total number of passengers is 36 (from the "Total" row and "Total" column).

Step2: Calculate probability

Probability is the number of favorable outcomes (front and window) divided by total outcomes. So, probability = $\frac{8}{36}$ = $\frac{2}{9}$. Wait, no, wait. Wait, the total number of passengers: let's check the table again. The "Total" column for "Total" is 36? Wait, the rows are Aisle, Window, Total? Wait, no, the columns are Front, Back, Total. Rows: Aisle, Window, Total. So total passengers: 14 (Aisle Total) + 22 (Window Total) = 36? Wait, no, the "Total" row: Front Total is 12, Back Total is 24, so total passengers is 12 + 24 = 36. The number of passengers who prefer front and window: looking at the "Window" row and "Front" column, that's 8. So probability is $\frac{8}{36}$? Wait, no, wait the question is "prefers to sit in the front of the plane and prefers a window seat". So front (column) and window (row). So the cell where Front column and Window row intersect is 8. Total passengers: 36 (12 + 24 or 14 + 22). So 8/36 simplifies to 2/9? Wait, but let's check the options. Option A is 2/9. Wait, but wait, maybe I misread the table. Wait, the table:

Columns: Front, Back, Total

Rows: Aisle, Window, Total

Aisle row: Front=4, Back=10, Total=14

Window row: Front=8, Back=14, Total=22

Total row: Front=12, Back=24, Total=36

So the number of passengers who prefer front (column) and window (row) is 8 (Front column, Window row). Total passengers is 36 (Total row, Total column). So probability is 8/36 = 2/9. So that's option A.

Answer:

A. $\frac{2}{9}$