Question

A flight attendant surveyed passengers on a flight about where they would prefer to sit. The results are shown in the Aisle WindowTotal 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?

Answer 1

The question of probabilities takes into account the number of required outcomes over the number of total outcomes.

A look at the table shows that the intersection of front and window seats is at 8. That is the first figure on the second column indicates those who would prefer a fron seat by the window. Note that there is a total of 36 passengers.

Hence the probability that a passenger prefers to sit in the front of the plane and prefers a window seat is shown as

$\begin{gathered} P(\text{front and window)=}\frac{Number\text{ of required outcomes}}{Number\text{ of possible outcomes}} \\ P(\text{front and windows)=}\frac{8}{36} \\ P(\text{fron and window)=}\frac{2}{9} \end{gathered}$

The correct answer is option D