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The seating chart of an airplane shown 30 rows of seats
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Updated on: 21 Oct 2018, 07:01
Updated on: 21 Oct 2018, 07:01
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62% (01:52) correct
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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)\(\frac{1}{6}\)
(B)\(\frac{1}{3}\)
(C)\(\frac{2}{3}\)
(D)\(\frac{5}{6}\)
(E)\(\frac{17}{18}\)
(A)\(\frac{1}{6}\)
(B)\(\frac{1}{3}\)
(C)\(\frac{2}{3}\)
(D)\(\frac{5}{6}\)
(E)\(\frac{17}{18}\)
Re: The seating chart of an airplane shown 30 rows of seats
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20 Oct 2018, 19:49
20 Oct 2018, 19:49
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sarahl 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)\(\frac{1}{6}\)
(B)\(\frac{1}{3}\)
(C)\(\frac{2}{3}\)
(D)\(\frac{5}{6}\)
(E)\(\frac{17}{18}\)
(A)\(\frac{1}{6}\)
(B)\(\frac{1}{3}\)
(C)\(\frac{2}{3}\)
(D)\(\frac{5}{6}\)
(E)\(\frac{17}{18}\)
Total seats = 30 *3 = 90
and total seat with obscured view = 5 * 3 = 15
Now total seat with unobscured view = 90 - 15 = 75
Therefore the probability that the person will get a seat with an unobscured view = \(\frac{75}{90} = \frac{5}{6}\).
*** Please at the end of your question do mention the source of the question***
Re: The seating chart of an airplane shown 30 rows of seats
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21 Oct 2018, 07:00
21 Oct 2018, 07:00
Expert Reply
sarahl 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)\(\frac{1}{6}\)
(B)\(\frac{1}{3}\)
(C)\(\frac{2}{3}\)
(D)\(\frac{5}{6}\)
(E)\(\frac{17}{18}\)
(A)\(\frac{1}{6}\)
(B)\(\frac{1}{3}\)
(C)\(\frac{2}{3}\)
(D)\(\frac{5}{6}\)
(E)\(\frac{17}{18}\)
hi, OA marked is wrong ...
It will be \(\frac{5}{6}\)....
There are 30 rows and each row has 6 seats, 3 on each side of centre aisle.. so total seats = 6*30 = 180
there are 2 window seats in each row, one on each side of aisle, so total = \(30*2=60\)
seats with obscured view is \(5*2=10\)..
So, the probability that the person will get a seat with an unobscured view = \(\frac{60-10}{60}=\frac{5}{6}\)
