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Joshua and Jose work at an auto repair center with 4 other

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Joshua and Jose work at an auto repair center with 4 other [#permalink] New post 27 Jun 2006, 21:20
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Jul 2014, 01:28, edited 1 time in total.
Edited the question and added the OA.
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post 27 Jun 2006, 23:28
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Two Methods

1) Probability of chosing Josh first = 1/6
Probability of chosing Jose second = 1/5
total = 1/30
Probability of chosing Jose first = 1/6
Probability of chosing Josh second = 1/5
Total = 1/30
Final = 1/30 + 1/30 = 1/15

2) Number of ways two persons can be chosen 6C2 = 15
Number of ways Josh and Jose are the two persons = 1
Total = 1/15
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post 13 Jul 2014, 19:23
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post 13 Jul 2014, 21:57
Total ways of choosing two people out of 6 = 6c2=15.
And out of these 15 cases there is only one case in which Joshua and Jose both are selected. hence 1/15 is the probability.

The other 14 cases would be : When Joshua is selected and one person out of the other four (excluding Jose) is selected. 1x4c1=4 cases. When Jose is selected and one person out of the other four (excluding Joshua) is selected. 1x4c1=4 cases. When none of Joshua and Jose is selected. 4c2=6 cases. This adds up to 4+4+6=14 cases.
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post 07 Sep 2014, 11:06
aiming700plus wrote:
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



I am not able to solve this questions with below method:

probability of not choosing both of them (probability of choose 2 from other members) = 4C2/6C2 = 6/15=2/5

probability required = 1-2/5 = 3/5

Where am I doing wrong?

Regards,
Ravi
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post 07 Sep 2014, 11:13
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email2vm wrote:
aiming700plus wrote:
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



I am not able to solve this questions with below method:

probability of not choosing both of them (probability of choose 2 from other members) = 4C2/6C2 = 6/15=2/5

probability required = 1-2/5 = 3/5

Where am I doing wrong?

Regards,
Ravi


You should also subtract committees with Joshua but not Jose and the committees with Jose but not Joshua.
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post 07 Sep 2014, 11:36
Bunuel wrote:
email2vm wrote:
aiming700plus wrote:
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



I am not able to solve this questions with below method:

probability of not choosing both of them (probability of choose 2 from other members) = 4C2/6C2 = 6/15=2/5

probability required = 1-2/5 = 3/5

Where am I doing wrong?

Regards,
Ravi


You should also subtract committees with Joshua but not Jose and the committees with Jose but not Joshua.



Hmmm...silly again.

Only Joshua but not jose probability = 4/15
only jose but not joshua = 4/15

2/5 +4/15+4/15= 14/15

so actual probabilty required is 1-14/15 = 1/15

Cheers Bunuel!!
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Joshua and Jose work at an auto repair center with 4 other [#permalink] New post 14 Oct 2014, 09:44
Bunuel

I went for a simpler approach and although I get the right answer, I'm not sure if it is correct. Here is how I went about it:

Since the question is only concerned about Josh and Jose, the probability that either one of them is chosen first is: 2/6
After the first is chosen, there are 5 people left, out if which one is either Josh or Jose. Probability of getting them is then: 1/5

So, (2/6)*(1/5)= 1/15

Is this a right approach for this question? To me this seems so much more straightforward...Thoughts? Suggestions?
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post 14 Oct 2014, 10:07
Expert's post
nyoebic wrote:
Bunuel

I went for a simpler approach and although I get the right answer, I'm not sure if it is correct. Here is how I went about it:

Since the question is only concerned about Josh and Jose, the probability that either one of them is chosen first is: 2/6
After the first is chosen, there are 5 people left, out if which one is either Josh or Jose. Probability of getting them is then: 1/5

So, (2/6)*(1/5)= 1/15

Is this a right approach for this question? To me this seems so much more straightforward...Thoughts? Suggestions?


Yes, that's correct.
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post 14 Oct 2014, 11:46
I am not able to solve this questions with below method:

probability of not choosing both of them (probability of choose 2 from other members) = 4C2/6C2 = 6/15=2/5

probability required = 1-2/5 = 3/5

Where am I doing wrong?

Regards,
Ravi[/quote]

You should also subtract committees with Joshua but not Jose and the committees with Jose but not Joshua.[/quote]


Hmmm...silly again.

Only Joshua but not jose probability = 4/15
only jose but not joshua = 4/15


2/5 +4/15+4/15= 14/15

so actual probabilty required is 1-14/15 = 1/15

Cheers Bunuel!![/quote]
I'm afraid I don't get the red part: only joshua and not jose: 1/6 * 4/5 = 4/30 =2/15. Do we now multiply by 2 because you can pick them in two ways (pick anyone but jose first and pick joshua second) ?
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post 15 Oct 2014, 10:49
usre123 wrote:
I am not able to solve this questions with below method:

probability of not choosing both of them (probability of choose 2 from other members) = 4C2/6C2 = 6/15=2/5

probability required = 1-2/5 = 3/5

Where am I doing wrong?

Regards,
Ravi


You should also subtract committees with Joshua but not Jose and the committees with Jose but not Joshua.[/quote]


Hmmm...silly again.

Only Joshua but not jose probability = 4/15
only jose but not joshua = 4/15


2/5 +4/15+4/15= 14/15

so actual probabilty required is 1-14/15 = 1/15

Cheers Bunuel!![/quote]
I'm afraid I don't get the red part: only joshua and not jose: 1/6 * 4/5 = 4/30 =2/15. Do we now multiply by 2 because you can pick them in two ways (pick anyone but jose first and pick joshua second) ?[/quote]

=====================
A> Only Joshua(Jh) and !Jose(Jo)

so we have Joshua + Any of the (A,B,C,D)= (Jh,A) or (Jh,B) or (Jh,C) or (Jh,D) ------>[or simply we can write it as 4c1 since we have Jh already and we have to choose one from other 4]

4c1/6c2

B> Only Jo and !Jh

same 4c1/6c2

final answer =

1- (probabilty of not choosing both of them)- probability of choosing Jh but !Jo - Probability of choosing Jo but !Jh.


This is bulky and I was just checking if things would work this way...simple way is

there can only be one set with both Jo and Jh and total number of ways choosing 2 from 6 is 6c2 = 15

therefore probability is 1/15 :-)
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post 15 Oct 2014, 12:57
email2vm wrote:
usre123 wrote:
I am not able to solve this questions with below method:

probability of not choosing both of them (probability of choose 2 from other members) = 4C2/6C2 = 6/15=2/5

probability required = 1-2/5 = 3/5

Where am I doing wrong?

Regards,
Ravi


You should also subtract committees with Joshua but not Jose and the committees with Jose but not Joshua.



Hmmm...silly again.

Only Joshua but not jose probability = 4/15
only jose but not joshua = 4/15


2/5 +4/15+4/15= 14/15

so actual probabilty required is 1-14/15 = 1/15

Cheers Bunuel!![/quote]
I'm afraid I don't get the red part: only joshua and not jose: 1/6 * 4/5 = 4/30 =2/15. Do we now multiply by 2 because you can pick them in two ways (pick anyone but jose first and pick joshua second) ?[/quote]

=====================
A> Only Joshua(Jh) and !Jose(Jo)

so we have Joshua + Any of the (A,B,C,D)= (Jh,A) or (Jh,B) or (Jh,C) or (Jh,D) ------>[or simply we can write it as 4c1 since we have Jh already and we have to choose one from other 4]

4c1/6c2

B> Only Jo and !Jh

same 4c1/6c2

final answer =

1- (probabilty of not choosing both of them)- probability of choosing Jh but !Jo - Probability of choosing Jo but !Jh.


This is bulky and I was just checking if things would work this way...simple way is

there can only be one set with both Jo and Jh and total number of ways choosing 2 from 6 is 6c2 = 15

therefore probability is 1/15 :-)[/quote]

Got it thanks. but what is wrong with my logic? if you could help, that'll be great. It'll help me conceptually.
Re: Joshua and Jose work at an auto repair center with 4 other   [#permalink] 15 Oct 2014, 12:57
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