The seating chart of an airplane shows 30 rows of seats

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.

Hey, i also got D.But my approach was different.Please correct me if i am wrong..

A person can be chosen from 30 window seats in 30C1 ways.
Now ,if we choose a person who sits in those 5 window seats which are obscured,we can do that in 5C1 ways
So probability(seating in a obscured seat)=5C1/30C1=5/30=1/6
Thus probability(seating in a non obscured seat)=1-1/6=5/6

+1 D.... Finally I got a probability sum correct

Hi Bunuel,

I've a doubt here "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."

So what I've understood is that in each row= 3 seats (Left, Center, Right), and ONE of them has window, (Left or Right only). So if each window has obscured view, ie. in total there are = 1 obscoured seat *5 rows= 5 seats with obscoured view.
There are total 90 seats, so 85 seats have unobscured view. Hence, 85/90.

Pls explain.

priyalr

6 seats per row, think of a boeing 737. We have 30 rows, therefore window 30 seats one one side and 30 window seats on the other, totaling 60 window seats on the whole plane.

the view of the window of 5 rows is blocked. two wings, therefore 10 window seats are blocked.

Total window seats = 60
Total blocked window seats = 10
Total unblocked seats = 50

We know that a window seat was given, therefore probability for not window seat is 50/60 =5/6

ANS DEEEEEE

Bunuel:

The question tells that each row has 3 seats on each side of the center aisle. As 3 on each side, total 6 seats and 30 rows.
therefore total seats would be 30*6 = 180
as 2 window seats in each row, we have 2*30 = 60 window seats
obscured view = 5*2 = 10
unobstructed view = 50

so, probability of getting an unobstructed window seat = 50/60 = 5/6

I'm not sure on how you got the number of total seats as 90. can you please explain.

The number of views on one side is 90. Because of the symmetry one can find the probability for 90 (one) and it will be the same for the 180 (total).

total seats in plane = 30*3 ; 90
seats with window ; 2*30 ; 60
window seats with obstruction ; 5*2 ; 10
window seats without obstruction ; 60-10 ; 50
so probability that the person will get a seat with an unobscured view ' 50/60 ; 5/6
IMO D

Bunuel:
I few take symmetry I end up with the following:

Total seats = 90
window seats = 30
5 are obscured, unobscured = 25
probability = unobscured / total window seats = 25/30 = 5/6.

I think the numbers you have in your answer do not match.

your answer is as follows:
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.

Am I missing something??

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

Given: 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.

Asked: 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?

Total number of window seats = 30*2 = 60
Number of window seats with obscured view = 5*2 = 10
Number of window seats with unobscured view = 60 - 10 = 50

The probability that the first person with get a seat with an unobscured view = 50/60 = 5/6

IMO D

Hi, Bunuel
how is # of seats with obscured view is 5*3=15

It's not...he just made a boo boo...shhhhh

30*2 = 60 total window seats.
(2*5) = 10 obscured window seats.
50 unobscured window seats.
50/60 = 5/6
Answer choice D.

stoned Nice signature...I respect it!!

Thanks, ThatDudeKnows. Bunuel doesn't even make a typo, that's why I asked.
About the signature..... I think this is the key to GMAT not emphasised enough.

I tried explaining this here. You could do it in even easier way: 5/30 = 1/6 of the seats have obscured view and the remaining 5/6 of the seats have unobscured view.

I was under the impression that PS questions only give you information you NEED to solve and are unlike DI questions in the sense they don't give you extra info to "bog you down". It seems pretty clear to me the point of this question is to do just that and throw you off by including unnecessary info.