Asif123 wrote:
The seating chart of an airplane shown 30 rows of seats. Each row has 3 seats on each side of the center aisle, and one of the seats on each side is a window seat. The view from the window seats in 5 of the rows is obscured by the wings of the airplane. If the first person to be assigned a seat is assigned a window seat and the window seat is assigned randomly, what is the probability that the person will get a seat with an unobscured view?
(A) 1/6
(B) 1/3
(C) 2/3
(D) 5/6
(E) 17/18
For some reason I got D as the answer.
Can someone point out the mistake in my logic:
We have 30 rows of seats. Each row consists of 6 seats divided by the aisle and each row have 2 windows seats.
So we have 60 window seats in total.
5 rows have obscured view window seats. So out of total 60 window seats we have 10 window seats with an obscured view.
If one is 100% assigned to window seat then the probability to get a seat with an obscured view is 10/60 = 1/6
Probability to get a seat with an unobscured view is 1 - 1/6 = 5/6