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

Question

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%

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

ScottTargetTestPrep · Accepted Answer

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:

- -

Since K and G are sitting next to each other, we can count   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

emockus · Answer

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%.

giobas · Answer

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%

carlo.damore1@gmail.com · Answer

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%

maxschmid · Answer

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.

arshiyakhanna · Answer

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%

stne · Answer

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%

Sreeragc · Answer

Can anyone please help in understanding the problem which i had some difficulties with. The problem says 4 random seats are on first come first serve basis in a row, it is only mentioned that K & G end up in same row and not said that in the same 4 random seats assigned on FCFS basis. So isn't there any probability for them to end up in seats other than the 4 seats. So they may come first and sits next in that 4 seats and may come and six next to one other in the remaining seats. This thought really caught me, can any please help me get out of this please.

stne · Answer

Hi  Sreeragc ,

Thank you for sharing your query:

May be this will help
Passengers are only assigned row numbers they are not assigned seat numbers.
K and G receive the  same  row number, hence they will be in the same row, doesn't matter which row they will be in, question is asking probability of them sitting next to each other, it's already given they will be in the same row.

Question is not asking , find probability of them sitting in the same row and then sit next to each other.
It's already given they are given the same row number.Don't think of choosing a row from various rows, that is not being asked.

Let me know if anything is still unclear.

Sreeragc · Answer

Thank you stne , that was a great help, four seats within the row was the real point i was missing. Now i got it, each row only has four sets. Now its good to go with the rest. Thanks again.

drsuryagoutham · Answer

Is the following method correct?

The four seats can be:
 A   B   C   D

The probability of Karen sitting in one of the seats is P(K) = 1/4
The probability of Georgia sitting in seat closest to Karen, from the remaining 3 is P(next to K) = 1/3

Karen and Georgia can be seated together in AB, BC or CD. So there are 3 possibilities here.

Also, Karen and Georgia could be seated interchangeably in the two seats. So there are another 2 possibilities here.

Overall, the total probability of them sitting together

= P(K) * P(next to K) * 3 * 2

= 1/4 * 1/3 * 3 * 2 = 6/12 = 50%

D4kshGargas · Answer

Hi,
Why are you assuming that the row has just 4 seats??

MBAB123 · Answer

It is given in the question - 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?

ScottTargetTestPrep · Answer

Response:

The question states “customers are assigned a row number, but the four seats within the row are first come, first served.” From this sentence, we understand that there are four seats in each row.

Bambi2021 · Answer

My thinking was like this:

For seats A-B-C-D, There is only 3 pairs of seats that will get them next to each other: AB, BC, CD.

In total though, you can choose two seats in 4C2 ways = 6.

3/6 = 50 %

CEdward · Answer

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%

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

Denominator: 
4 x 3 x 2 x 1 = 24 <--- Total number of possibilities

Numerator: 
3 x 2 x 1 = 6 <--- 1 possible arrangement in which K and G sit next to each other, but they can swap places ...

So ...6 x 2 = 12

12 / 24 = 50%

C.

Scuven · Answer

Suppose a 4 seats series in a row.

Here not only combination, but also arrangements count.

Thus we use FCF to assign these 2 people to 4 seats (counting also cases in which they are arrange thus that K-G and G-K are two cases)

Total cases= 4*3=12 total ways in which K and G can sit with and without being together.

Then assume that Karen and G are together and thus must be counted as 1 person. We have 3 seats to arrange.

Thus there are 3*2= 6 arrangements with K and G together

6/12=0,5

Ved22 · Answer

No need for such calculations. Simply use the concept of symmetrical. In these 4 seats, in half that is 50% are the possibilities that Both are together. If you are aware of the symmetric concept then it is a 30 secs sum

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