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The seating chart of an airplane shows 30 rows of seats

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The seating chart of an airplane shows 30 rows of seats  [#permalink]

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New post 24 Feb 2012, 22:09
2
2
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A
B
C
D
E

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  75% (hard)

Question Stats:

53% (01:20) correct 47% (01:48) wrong based on 137 sessions

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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
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Re: PT # 8 PS 2 Q 19  [#permalink]

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New post 24 Feb 2012, 22:20
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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.
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Re: The seating chart of an airplane shows 30 rows of seats  [#permalink]

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New post 24 Mar 2012, 22:38
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
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Re: The seating chart of an airplane shows 30 rows of seats  [#permalink]

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New post 01 Apr 2012, 03:26
Pls correct if m wrong

Total number of window seats 30+30 = 60(on both sides of the aisle
Obscured on both sides 5+5 =10
Unobscured on both sides =50

Probability = unobscured /total window seats
50/60 = 5/6
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Re: The seating chart of an airplane shows 30 rows of seats  [#permalink]

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New post 01 Apr 2012, 04:22
+1 D.... Finally I got a probability sum correct :-P
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Re: The seating chart of an airplane shows 30 rows of seats  [#permalink]

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New post 01 Apr 2012, 15:09
How come you didn't consider 6 seats per row? Isn't that what is stated in the question? I know the numbers workout the same but still.

# of Total seats = 30(3)(2) = 180

Window seats = 60
Obstructed view window seat = 10

Since the question tells us that a window seat was obtained, all we care about are the window seats

Therefore probability of unobstructed view = 60-10/60 = 5/6 D

Is this reason of thinking correct? or the above is better?
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Re: The seating chart of an airplane shows 30 rows of seats  [#permalink]

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New post 04 Apr 2012, 07:36
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.
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Re: The seating chart of an airplane shows 30 rows of seats  [#permalink]

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New post 04 Apr 2012, 11:34
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

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Re: The seating chart of an airplane shows 30 rows of seats  [#permalink]

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