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its an easy question

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its an easy question

by sana.noor » Tue Sep 24, 2013 10:12 pm
its an easy question which can be done in 2 ways. i can get answer in one way but couldn't find the right answer when going through the other way

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%
OA is B
if we adjust bob and lisa to the two ends of the row then the other 3 people can adjust in 3!= 6 ways. now bob and lisa can swap the two end thus 6.2= in 12 ways they both can sit with other group members. the total outcomes are 5! = 120 and thus the probability will be 12/120 = 10%

what if we let the two sit together and then subtracting it from the total possible outcomes to find the probability? i am not getting answer that way
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by ganeshrkamath » Tue Sep 24, 2013 11:18 pm
sana.noor wrote:its an easy question which can be done in 2 ways. i can get answer in one way but couldn't find the right answer when going through the other way

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%
OA is B
if we adjust bob and lisa to the two ends of the row then the other 3 people can adjust in 3!= 6 ways. now bob and lisa can swap the two end thus 6.2= in 12 ways they both can sit with other group members. the total outcomes are 5! = 120 and thus the probability will be 12/120 = 10%

what if we let the two sit together and then subtracting it from the total possible outcomes to find the probability? i am not getting answer that way
Favourable case: B _ _ _ L and L _ _ _ B
= 2 * 3! = 12

Sample space = 5! = 120

Probability = 12/120 = 0.1 or 10%

Choose [spoiler](b)[/spoiler]

Bob and Lisa sitting together doesn't cover all the cases.
You will only cover :
BL _ _ _
_ BL _ _
_ _ BL _
_ _ _ BL
You would've missed :
B _ L _ _
B _ _ L _
and so on.


Hope this helps.
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by theCodeToGMAT » Tue Sep 24, 2013 11:22 pm
The Other method is

TOTAL WAYS - (WAYS IN WHICH BOTH SITH WITH TWO OTHERS) - (ONE TAKES CORNER SEAT AND OTHER SITS WITH BOTH)
= 5! - 2!(2c2 x 3c1)3! - 2!(1c1x3c2)3!x2!
= 120 - 2x1x3x6 - 2x1x3x6x2
= 120 - 36 - 72
= 12

hence, 12/120*100 = 10%

For your Info:
2!(2c2 x 3c1)3! --> B & L takes 3 middle seats so 2C2 and 3C1 .... they can be arranged so 3! .. the corner seats can also be arranged so 2!


2!(1c1x3c2)3!x2! --> B & L either can take corner seat so 2!... out of 4 other seats B/L will take one so 1C1.. other 3 will take 2 centre positions.. so 3c2 .. they can arrange themselves.. so 3!... this scenarios can happen on either corner.. so again 2!..
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by vipulgoyal » Wed Sep 25, 2013 1:42 am
Hi Sana, here we go

1 - not desired
not desired cases = when Bob and Lina both sit in the middle + when Lina sit in the middle and Bob sit in the extrem + when Bob sit in the middle and Lina sit in the extrem

case 1. - - - - - , out of middle three places two places are fixed for bob and lina and third place is for any one out of three, 3c1*3*2*1*2/5! = 3/10------------- 1

case 2 and 3. out of lina and bob one will sit at the middle three places and another one will sit at either extrems

2*3c2*3*2*1*2/5! = 6/10 ---------------------------- 2

how 2*3c2*3*2*1*2/5!

for middle three positions one is for either bob or lina , 2 ways, select 2 out of three other thn lina and bob 3c2, now these three can be arranged by 3! in middle three positions so 3*2*1, now the extrem two positions can be filled by two differant ways

1 - 3/10 + 6/10 = 1/10
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by Brent@GMATPrepNow » Wed Sep 25, 2013 5:27 am
sana.noor wrote: 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%
Here's a solution using probability rules.

First, we must recognize that Bob and Lisa must sit on the extreme ends (seats 1 and 5 in the row).

So, we can reword the question as "What is the probability that Bob and Lisa both sit on the extreme ends?"

P(both sit on extreme ends) = P(one of them sits on left-most seat AND the other sits on the right-most seat)

= P(one of them sits on left-most seat) X P(the other sits on the right-most seat)

= (2/5) X (1/4)
= [spoiler]1/10[/spoiler]
= B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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