Michele4 wrote:
Bunuel wrote:
eybrj2 wrote:
The seating chart of an airplane shows 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 saet. 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 thw 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
Total # of seats is 30*3=90;
# of seats with obscured view is 5*3=15, hence # of seat with unobscured view is 90-15=75;
P=favorable/total=75/90=5/6.
Answer: D.
Bunuel chetan2u GMATNinja why the number of seats with obscured view is 5*3 if we got 1 seat per side?
S CCC S
Hi,
Aisle means a passage.
The seats are W, S, S......S, S, W , where W shows window seats, S represents normal seats and '...' shows the aisle.
So each row has 6 seats, out of which 2 are window.
Since the question talks of only window seats, the total Ws are 30*2=60.
But 5 rows have obstructed view, so 5*2=10 have obstructed view.
Therefore P of the seat not having obstructed view = \(\frac{60-10}{60}=\frac{50}{60}=\frac{5}{6}\)
D