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On a certain airline, customers are assigned a row number when they pu

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On a certain airline, customers are assigned a row number when they pu [#permalink]

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New post 06 Mar 2017, 03:56
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Question Stats:

55% (01:20) correct 45% (01:03) wrong based on 96 sessions

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On a certain airline, customers are assigned a row number when they purchase their ticket, but the four seats within the row are first come, first served during boarding. If Karen and Georgia end up with random seats in the same row on a sold-out flight, what is the probability that they sit next to each other?

A. 25%
B. 40%
C. 50%
D. 75%
E. 80%

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Re: On a certain airline, customers are assigned a row number when they pu [#permalink]

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New post 06 Mar 2017, 07:03
i dont know if i am correct but here how i would do it:
probability of karen sitting on the edge of the row is 2/4 = 1/2 , if karen sits on the edge there is one possibility for georgia to sit besides her so probability would be 1/3 for here. together 1/2*1/3=1/6
the probability of karen sitting mid is 1/2 and probability of georgia sitting besides her is 2/3 - so together 1/2*2/3=2/6

together 1/6 + 2/6 = 3/6 = 50%
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On a certain airline, customers are assigned a row number when they pu [#permalink]

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New post Updated on: 06 Mar 2017, 08:54
Easiest way to solve, for myself atleast, is to illustrate: let "x" be the customers and "o" be the open seats

x-x-o-o
o-x-x-o
o-o-x-x
x-o-x-o
o-x-o-x
x-o-o-x

3 ways that the pair can sit together out of a total of 6 arrangements. Simply divide and the result is 50%.
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Originally posted by emockus on 06 Mar 2017, 07:35.
Last edited by emockus on 06 Mar 2017, 08:54, edited 2 times in total.
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Re: On a certain airline, customers are assigned a row number when they pu [#permalink]

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New post 06 Mar 2017, 08:22
There is 4C2 ways Karen and Georgia can sit in the row.
There is 3C1 ways Karen and Georgia can sit togheter in the row (Imagine Karen and Georgia as a single unit, so there are just 3 other seats to ocuppy).

(Farovable outcomes)/(total outcomes) = 3C1/4C2 = 3/6 = 50%
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Re: On a certain airline, customers are assigned a row number when they pu [#permalink]

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New post 07 Mar 2017, 17:19
Bunuel wrote:
On a certain airline, customers are assigned a row number when they purchase their ticket, but the four seats within the row are first come, first served during boarding. If Karen and Georgia end up with random seats in the same row on a sold-out flight, what is the probability that they sit next to each other?

A. 25%
B. 40%
C. 50%
D. 75%
E. 80%


In a row of 4 seats, there are 4! = 24 sitting arrangements for 4 people. Now let’s find the number of ways that Karen (K) and Georgia (G) will be sitting next to each other. Let’s designate the other two passengers as A and B, and they are sitting in the same row as Karen and Georgia. The arrangement could be as follows:

[KG]-[A]-[B]

Since K and G are sitting next to each other, we can count [KG] as one person, such that there are 3 people sitting in the same row, and thus the number of ways to arrange those 3 people is 3! = 6. However, since we can arrange KG in 2! ways (KG or GK), the number of ways to arrange the “3 people” is actually 2 x 6 = 12. Thus, the probability that Karen and Georgia are sitting next to each other is 12/24 = 1/2 = 50%.

Answer: C
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Re: On a certain airline, customers are assigned a row number when they pu [#permalink]

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New post 10 Mar 2017, 04:50
How can there be 12 different ways that they sit next to each other?
I even drew it out and still dont get it.

1 / 2 / 3 / 4
k / g / x /x
g/ k / x /x
x / k / g / x
x/ g / k / x
x / x / k / g
x / x / g / k

for me there would be 6 different ways. out of 24 possible arrangements.
therefore I would say answer A.

Please tell me where my mistake lies.
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Re: On a certain airline, customers are assigned a row number when they pu [#permalink]

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New post 10 Mar 2017, 09:27
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Consider K & G to be one group as they will be sitting next to each other. Therefore, the total arrangements among 3 seats (K&G is one & other two) will be 3!x2!, since within the group, K & G can sit in two ways. Divide by total no of possibilities - 4!. Therefore ans is 50%
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On a certain airline, customers are assigned a row number when they pu [#permalink]

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New post 20 Mar 2018, 06:46
Bunuel wrote:
On a certain airline, customers are assigned a row number when they purchase their ticket, but the four seats within the row are first come, first served during boarding. If Karen and Georgia end up with random seats in the same row on a sold-out flight, what is the probability that they sit next to each other?

A. 25%
B. 40%
C. 50%
D. 75%
E. 80%



Consider KG as one unit hence A{KG}B where A and B are the other two passengers hence total ways to arrange them is \(3_C_1\) = 3
A and B can be arranged in \(2_C_
{1}\) =2 ways
KG themselves can be arranged in \(2_C_{1}\) =2 ways
Hence total favorable cases that they can be arranged in are 3*2*2 =12
Total arrangements are 4*3*2*1 =24
hence probability = \(\frac{12}{24}\) = 50%
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On a certain airline, customers are assigned a row number when they pu   [#permalink] 20 Mar 2018, 06:46
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