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A flight attendant surveyed passengers on a flight about their seating preferences. The results are shown in the table below:

| | Aisle | Window | Total |
|--------|-------|--------|-------|
| Front | 4 | 8 | 12 |
| Back | 10 | 14 | 24 |
| Total | 14 | 22 | 36 |

What is the probability that a passenger prefers to sit in the front of the plane and prefers a window seat?

A. [tex]\( \frac{2}{9} \)[/tex]
B. [tex]\( \frac{1}{3} \)[/tex]
C. [tex]\( \frac{4}{11} \)[/tex]
D. [tex]\( \frac{2}{3} \)[/tex]


Sagot :

To solve the problem, we need to determine the probability that a passenger prefers to sit in the front of the plane and prefers a window seat. Here's the step-by-step solution:

1. Identify the number of passengers who prefer both the front and a window seat:
According to the given table, the number of passengers who prefer to sit in the front and prefer a window seat is 8.

2. Find the total number of surveyed passengers:
The total number of passengers surveyed is 36, as given in the table.

3. Calculate the probability:
The probability of an event is given by the ratio of the favorable outcomes to the total outcomes. In this case, the probability [tex]\( P \)[/tex] that a passenger prefers to sit in the front of the plane and prefers a window seat is:
[tex]\[ P = \frac{\text{Number of passengers preferring front and window seat}}{\text{Total number of passengers}} \][/tex]
Substituting the values, we get:
[tex]\[ P = \frac{8}{36} = \frac{2}{9} \][/tex]

Therefore, the correct answer is:
A. [tex]\(\frac{2}{9}\)[/tex]
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