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Joshua and Jose

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SEPY
 Post subject: Joshua and Jose
 Post Posted: Fri Sep 17, 2010 10:51 am 
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Students


Posts: 12
Joshua and Jose work at an auto repair center with 4 other workers. For a survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will both be chosen?

a)1/15
b)1/12
c)1/9
d)1/6
e)1/3

Hi Ron,

I have a doubt here with respect to the method where order matters or where it does not.

I was trying to solve this problem by using Combinatorics strategy.(Slot method)
I got total cases by that strategy as (6X5)/2! = 15
But how Joshua and Jose can be selected in only one way?
I have concluded by approach : 1 1
and as order does not matter between Joshua and Jose selection,therefore divided (1X1) by 2!(which is wrong :-(

Please help me to understand this..

Answer to this is 1/15
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gokul_nair1984
 Post subject: Re: Joshua and Jose
 Post Posted: Fri Sep 17, 2010 11:32 am 
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Students


Posts: 170
We can use probability:
The probability of selecting Joshua and Jose from 6 people( both inclusive) can be given by:

P(Joshua)*P(Jose)+P(Jose)*P(Joshua)
ie; Probability of first selecting Joshua and Probability of then selecting Jose or Probability of first selecting Jose and Probability of then selecting Joshua .
ie; (1/6)*(1/5)+(1/6)*(1/5)=2/30=1/15


dipti.ch12 wrote:
I have a doubt here with respect to the method where order matters or where it does not.

I was trying to solve this problem by using Combinatorics strategy.(Slot method)
I got total cases by that strategy as (6X5)/2! = 15
But how Joshua and Jose can be selected in only one way?
I have concluded by approach : 1 1
and as order does not matter between Joshua and Jose selection,therefore divided (1X1) by 2!(which is wrong :-(



or else you could consider 2 slots ____ ____
For the first slot, any of the 6 people can be selected and for the second one any of the remaining 5 can be selected. Thus 6*5=30.
Now we need (2 specific people) to actually occupy the slots. Thus the first slot could be occupied by any of the 2(Joshua or Jose) in 2 ways. The remaining slot has only 1 selection remaining.2*1=2
Prob=2/30=1/15

or selecting 1 ad-hoc pair out of the total number of pairs(6C2=15)) =>1/15

Hope it is clear.
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SEPY
 Post subject: Re: Joshua and Jose
 Post Posted: Fri Sep 17, 2010 11:57 am 
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Students


Posts: 12
Hi Gokul,

Thanks for detailed explanation.
Probability technique is clear to me.I wanted to apply the concept told by Ron in one of his workshop"Thursday's with Ron"(3 dec 2009)
I am able to solve almost every question which pertains to combinations or Prob with that..But this one I am goofing up somewhere :-(
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tim
 Post subject: Re: Joshua and Jose
 Post Posted: Sat Sep 25, 2010 6:54 am 
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ManhattanGMAT Staff


Posts: 2242
Location: Southwest Airlines, seat 21C
Your answer already DID divide by 2! - notice you wrote a 2! yourself when you came up with 15 choices. Using combinatorics, there are 15 ways to choose 2 from a group of 6, as you indicate. How many ways are there to choose 2 from a group of 2? Only one: you pick both of them. Remember order doesn't matter here, so you don't have 2 choices, just 1. Take that 1 out of 15 and you have the correct answer..

_________________
Tim Sanders
Manhattan GMAT Instructor
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