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A group of 5 students bought movie tickets in one row next [#permalink]
20 Jun 2007, 16:03

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B

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Question Stats:

53% (02:21) correct
47% (04:15) wrong based on 298 sessions

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 Bob and Lisa will each sit next to only one of the four other students from the group?

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

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...

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%

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.

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.

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.

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.

Re: A group of 5 students bought movie tickets in one row next [#permalink]
09 Sep 2013, 01:59

3

This post received KUDOS

Expert's post

chetan86 wrote:

Bunuel,

Could you pls set OA and timer to this question?

Done.

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 Bob and Lisa will each sit next to only one of the four other students from the group? (A) 5% (B) 10% (C) 15% (D) 20% (E) 25%

The question basically asks about the probability that Bob and Lisa sit at the ends.

The total # of sitting arrangements is 5!.

Desired arrangement is either BXYZL or LXYZB. Now, XYZ can be arranged in 3! ways, therefore total # of favorable arrangements is 2*3!.

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%

Re: A group of 5 students bought movie tickets in one row next [#permalink]
20 Feb 2015, 06:35

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Re: A group of 5 students bought movie tickets in one row next [#permalink]
20 Feb 2015, 12:41

1

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Expert's post

Hi All,

There are several different 'math' approaches to this question. You have to be clear that there are 2 'acceptable' orientations of the 5 people though:

Bob, X, X, X, Lisa Lisa, X, X, X, Bob

Since either Bob OR Lisa would have to be in the 'first spot', there's a 2/5 probability of that happening....

That would leave the other person to occupy the 'fifth spot'; since there are 4 people remaining, there's a 1/4 probability of that happening....

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...