|
Author |
Message |
|
TAGS:
|
|
|
VP
Joined: 10 Jun 2007
Posts: 1478
Followers: 5
Kudos [?]:
70
[0], given: 0
|
A group of 5 students bought movie tickets in one row next [#permalink]
 20 Jun 2007, 17:03
10. A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that both of them will sit next to only one other student from the group.
A. 5%
B. 10%
C. 15%
D. 20%
E. 25%
|
|
|
|
|
|
|
Manager
Joined: 11 Mar 2007
Posts: 70
Followers: 1
Kudos [?]:
2
[0], given: 0
|
I may be oversimplifying this, but isn't it just asking "what is the probability that Bob and Lisa are sitting on the outside of the group?"
1/5 * 1/4 = 1/20 = 5%
Disclaimer: I really suck at these problems, so I'm probably wrong.
|
|
|
|
|
|
VP
Joined: 10 Jun 2007
Posts: 1478
Followers: 5
Kudos [?]:
70
[0], given: 0
|
mavery wrote: I may be oversimplifying this, but isn't it just asking "what is the probability that Bob and Lisa are sitting on the outside of the group?"
1/5 * 1/4 = 1/20 = 5%
Disclaimer: I really suck at these problems, so I'm probably wrong.
I am not that good either. However, if they both are sitting outside, which I think the question is asking, the probability would be:
Bob, Lisa, x, x, x = 6 ways
Lisa, Bob, x , x, x = 6 ways
x,x,x Bob, Lisa = 6 ways
x,x,x, Lisa, Bob = 6 ways
So sum up to 24 ways
Total possible = 5! = 120
so P = 24 / 120 = 20%
This is incorrect though... I don't know what I did wrong.
OA: B
|
|
|
|
|
|
Manager
Joined: 11 Mar 2007
Posts: 70
Followers: 1
Kudos [?]:
2
[0], given: 0
|
Ok, you're thinking differently that I am...I was thinking
Bob,x,x,x,Lisa
or Lisa,x,x,x,Bob
So I may be stretching it a bit here, but maybe the way I calculated was for bob in the first seat and lisa in the 5th seat...maybe we need to multiply by 2 to get both scenarios? (bob in 1st, lisa in 5th - bob in 5th, lisa in 1st)
This makes sense ur way too, because it will cut your scenarios from 24 to 12...
|
|
|
|
|
|
VP
Joined: 10 Jun 2007
Posts: 1478
Followers: 5
Kudos [?]:
70
[0], given: 0
|
mavery wrote: Ok, you're thinking differently that I am...I was thinking
Bob,x,x,x,Lisa
or Lisa,x,x,x,Bob
So I may be stretching it a bit here, but maybe the way I calculated was for bob in the first seat and lisa in the 5th seat...maybe we need to multiply by 2 to get both scenarios? (bob in 1st, lisa in 5th - bob in 5th, lisa in 1st)
This makes sense ur way too, because it will cut your scenarios from 24 to 12...
Actually, if I use your intrepretation of the question, they answer is correct.
Say:
Bob, x, x, x, Lisa = 3! = 6 ways
Lisa, x, x, x, Bob = 3! = 6 ways
So P = 12 / 120 = 10%
I need to intrepret the question better 
|
|
|
|
|
|
Director
Joined: 30 Nov 2006
Posts: 598
Location: Kuwait
Followers: 8
Kudos [?]:
115
[0], given: 0
|
Here are the cases:
BXXXL - LXXXB - XLBXX - XBLXX - XXBLX - XXLBX
3! for each case
3! x 6 = 36
Total number of ways to arrange the 5 students accross the row is 5! = 5x4x3x2 = 120
Probability = 36/120 = 6/20 = 30/100 = 30 %
???!!!!
|
|
|
|
|
|
VP
Joined: 10 Jun 2007
Posts: 1478
Followers: 5
Kudos [?]:
70
[0], given: 0
|
Mishari wrote: Here are the cases:
BXXXL - LXXXB - XLBXX - XBLXX - XXBLX - XXLBX
3! for each case
3! x 6 = 36
Total number of ways to arrange the 5 students accross the row is 5! = 5x4x3x2 = 120
Probability = 36/120 = 6/20 = 30/100 = 30 %
???!!!!
The question said "both of them will sit next to only one other student"
I guess, you cannot have arrangement such as XLBXX because L & B are sitting next to two two students.
|
|
|
|
|
|
Director
Joined: 30 Nov 2006
Posts: 598
Location: Kuwait
Followers: 8
Kudos [?]:
115
[0], given: 0
|
What's the source of the question ?
|
|
|
|
|
|
VP
Joined: 10 Jun 2007
Posts: 1478
Followers: 5
Kudos [?]:
70
[0], given: 0
|
Mishari wrote: What's the source of the question ?
I don't know...it was emailed to me by another GMAT struggler...
|
|
|
|
|
|
Senior Manager
Joined: 03 Jun 2007
Posts: 386
Followers: 2
Kudos [?]:
8
[0], given: 0
|
10% is what I get too. I guess Lisa and Bob sit at the end
|
|
|
|
|
|
VP
Joined: 08 Jun 2005
Posts: 1172
Followers: 5
Kudos [?]:
78
[0], given: 0
|
I know that OA is (B) but I'm getting (D)
the only options according to the stem are:
BLXXX = 1/5*1/4 = 1/20 (chances of Bob sitting first and Lisa sitting next to him)
LBXXX = 1/5*1/4 = 1/20
XXXBL = 1/5*1/4 = 1/20
XXXLB = 1/5*1/4 = 1/20
total = 1/20+1/20+1/20+1/20 = 4/20 = 1/5 = 20%
the only way to get (B) is if the stem means:
LXXXB = 1/5*1/4 = 1/20
BXXXL = 1/5*1/4 = 1/20
1/20+1/20 = 2/20 = 10%
is this "both of them" or "each one of them" ???
|
|
|
|
|
|
Director
Joined: 12 Jun 2006
Posts: 544
Followers: 1
Kudos [?]:
6
[0], given: 1
|
Bob and Lisa are students too. Given the above conditions, there are 2 possible seating arrangments:
scenario 1: LXXXB or
scenario 2: BXXXL
there are 5(4)3(2)1 = 120 total seating arrangments. in scenario 1, bob must be seated on the far right and lisa on the far left. this leaves 6 different possible seating arrangements among the other 3 students: 3(2)1 = 6. 6/120 = 1/20.
the prob of scenario 2 happening is also 1/20. either scenario 1 or 2 will occur so we sum the probs of 1 or 2 happening, 1/20 + 1/20 + 2/20 = 1/10 = .10.
|
|
|
|
|
|
Director
Joined: 12 Jun 2006
Posts: 544
Followers: 1
Kudos [?]:
6
[0], given: 1
|
killersquirrel,
will you explain what's going on here:
Quote: LXXXB = 1/5*1/4 = 1/20
BXXXL = 1/5*1/4 = 1/20
where did 1/5*1/4 come from? it seems like you've used some sort of abbreviated method.
|
|
|
|
|
|
VP
Joined: 08 Jun 2005
Posts: 1172
Followers: 5
Kudos [?]:
78
[0], given: 0
|
ggarr wrote: killersquirrel, will you explain what's going on here: Quote: LXXXB = 1/5*1/4 = 1/20
BXXXL = 1/5*1/4 = 1/20 where did 1/5*1/4 come from? it seems like you've used some sort of abbreviated method. hello ggarr
If we have five people sitting in a raw (n=5) then the chance of Lisa sitting in the extreme left is 1/5.
since Lisa is sitting in the extreme left, we are left with 4 places for Bob to sit in.
The chance for Bob to sit in the extreme right is now 1/4 since Lisa occupy one spot.
Since the events are independent I multiply them - this process then repeated for Lisa sitting in the extreme right and Bob sitting in the extreme left.
The same way I would calculate in a coin toss
then we just need to combine both outcomes.
|
|
|
|
|
|
Manager
Joined: 12 Apr 2007
Posts: 171
Followers: 1
Kudos [?]:
1
[0], given: 0
|
this is a good one...
i at first thought that it meant by one OTHER student that they could sit next to each other.
ie XXBLX and XXLBX counted as favorable outcomes.
but if you interpret the question to mean that bob and lisa can only sit on the ends (which would have been a clearer way to ask the question) the OA makes sense.
5! = 120 outcomes
BXXXL * 3! = 6 ways to arrange the other 3 students in the middle.
and
LXXXB * 3! = 6 ways to arrange the other 3 students in the middle.
(6+6)/120 = 10%
|
|
|
|
|
|
Manager
Joined: 18 Apr 2007
Posts: 62
Followers: 1
Kudos [?]:
0
[0], given: 0
|
first find the total possible ways.
5 spots.. so 5x4x3x2x1
then put L and B at an end.. 3 spots remaining... 3x2x1
but B and L can swap spots at the end.. so u multiply that by 2=12
so 120/12=10%!!!
|
|
|
|
|
|
Director
Joined: 13 Mar 2007
Posts: 555
Schools: MIT Sloan
Followers: 4
Kudos [?]:
3
[0], given: 0
|
3! x 2! = 12
total ways = 5! = 120
12/120 = 10%
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
