Joshua and Jose work at an auto repair center with 4 other : GMAT Problem Solving (PS)
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 09 Dec 2016, 18:24

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 11 Jun 2006
Posts: 15
Followers: 0

Kudos [?]: 8 [1] , given: 0

Joshua and Jose work at an auto repair center with 4 other [#permalink]

### Show Tags

27 Jun 2006, 21:20
1
KUDOS
7
This post was
BOOKMARKED
00:00

Difficulty:

15% (low)

Question Stats:

73% (01:51) correct 27% (01:10) wrong based on 367 sessions

### HideShow timer Statistics

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.
SVP
Joined: 30 Mar 2006
Posts: 1737
Followers: 1

Kudos [?]: 76 [4] , given: 0

Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]

### Show Tags

27 Jun 2006, 23:28
4
KUDOS
3
This post was
BOOKMARKED
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
Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 408
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB
Followers: 8

Kudos [?]: 195 [0], given: 50

Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink]

### Show Tags

16 Aug 2010, 17:05
Safiya wrote:
I suck in probability questions! Could someone please explain me the solution?

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

(A) 1\15
(B) 1\12
(C) 1\9
(D) 1\6
(E) 1\3

We should consider Joshua and Jose as one entity and then proceed for the probability.

Total ways of choosing 2 people (Joshua and Jose) out of 6 people --
$$(6!)/(4! * 2!)$$
$$=15$$

Favorable outcomes - 1 (Since Joshua and Josh should be picked up)

Probability - $$1/15$$

Any other thoughts.....
_________________

Support GMAT Club by putting a GMAT Club badge on your blog

Director
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 17 Jul 2010
Posts: 690
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Followers: 15

Kudos [?]: 147 [0], given: 15

Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink]

### Show Tags

16 Aug 2010, 17:45
Interestingly in such questions if one notices the denominator, we know it has to be 15, as 6C2=15, so that rules out B C D. Leaves us with A and B. A says 1/15, B says 5/15. of course there is one choice and that is 1/15, but even if you did not get it the right way, this elimination should help.
_________________

Consider kudos, they are good for health

Manager
Joined: 06 Apr 2010
Posts: 58
Followers: 0

Kudos [?]: 29 [3] , given: 13

Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink]

### Show Tags

16 Aug 2010, 21:25
3
KUDOS
1
This post was
BOOKMARKED
I reworded this problem to have Josh and Jose be two guys, and the other 4 are women. What are the chances of picking only guys?

There are two slots, first slot has (2/6) chance of picking a guy, the second slot has (1/5) chance of picking the other guy.

Picture them in a group. The first time you randomly pick someone, there are six people, and two guys. Now, assuming you have picked a guy, you are left with a group of 5 people standing there (1 guy, 4 girls). The chance of picking a guy then is 1/5. Multiply the two probabilities together, and you have 2/30, or 1/15
_________________

If you liked my post, please consider thanking me with Kudos! I really appreciate it!

Intern
Joined: 03 Aug 2010
Posts: 13
Followers: 0

Kudos [?]: 11 [0], given: 0

Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink]

### Show Tags

17 Aug 2010, 01:23
Probability of an event = (number of positive outcomes)/(total number of outcomes)

Here number of positive outcomes = choose Josh and Jose when two people are picked at random = 2C2

Total Outcomes = choose two people from a group of 6 people = 6C2

Probability = 2C2 / 6C2 = 1/15
Senior Manager
Status: Fighting on
Joined: 14 Mar 2010
Posts: 318
Schools: UCLA (R1 interview-WL), UNC(R2--interview-ding) Oxford(R2-Admit), Kelley (R2- Admit ), McCombs(R2)
WE 1: SE - 1
WE 2: Engineer - 3
Followers: 4

Kudos [?]: 28 [0], given: 3

Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink]

### Show Tags

17 Aug 2010, 15:59
total ways to select 2 people out of 6 = 6c2 = 15
the number of ways to select J & J = 2c2 = 1

Senior Manager
Joined: 03 Nov 2005
Posts: 395
Location: Chicago, IL
Followers: 3

Kudos [?]: 48 [0], given: 17

Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink]

### Show Tags

17 Aug 2010, 16:28
P (picking Joshua) first is 1/6 and picking Ann second 1/5.

1 st case So, P(J & A)=1/5 x 1/6 = 1/30 - J. and Ann second.

and in the reverse order,

2nd case P(A & J) = 1/30- Ann first, Joshua second

_________________

Hard work is the main determinant of success

Manager
Joined: 17 Oct 2008
Posts: 196
Followers: 1

Kudos [?]: 23 [0], given: 11

Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink]

### Show Tags

17 Aug 2010, 20:14
rlevochkin wrote:
P (picking Joshua) first is 1/6 and picking Ann second 1/5.

1 st case So, P(J & A)=1/5 x 1/6 = 1/30 - J. and Ann second.

and in the reverse order,

2nd case P(A & J) = 1/30- Ann first, Joshua second

same method as yours.
1/15
Manager
Joined: 17 Oct 2008
Posts: 196
Followers: 1

Kudos [?]: 23 [0], given: 11

Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink]

### Show Tags

17 Aug 2010, 20:17
Can someone spot me the error in this alternate method,

total outcome: 6c2
non favourable outcome:4c2 ( leaving out josha and jose)

req prob -> 1- (4c2/6c2)
reduces to 9/15.!!
Intern
Joined: 15 Aug 2010
Posts: 23
Location: Mumbai
Schools: Class of 2008, IIM Ahmedabad
Followers: 5

Kudos [?]: 20 [0], given: 0

Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink]

### Show Tags

17 Aug 2010, 20:40
Hi,

Non Fav o/comes =
Leaving out Joshua and Jose +
Leaving out Joshua and including Jose +
Leaving out Jose and including Joshua

Leaving out Joshua and Jose = 4C2
Leaving out Joshua and including Jose = 4C1 (as Jose has already been selected)
Leaving out Jose and including Joshua = 4C1 (as Joshua has already been selected)

When these cases are eliminated, one does get the answer of 1/15.

Hope this helps. Thanks.
_________________

Naveenan Ramachandran
4GMAT - Mumbai

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12904
Followers: 562

Kudos [?]: 158 [0], given: 0

Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]

### Show Tags

13 Jul 2014, 19:23
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Intern
Joined: 13 Aug 2013
Posts: 23
Followers: 0

Kudos [?]: 4 [0], given: 1

Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]

### Show Tags

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.
Manager
Status: folding sleeves up
Joined: 26 Apr 2013
Posts: 157
Location: India
Concentration: Finance, Strategy
GMAT 1: 530 Q39 V23
GMAT 2: 560 Q42 V26
GPA: 3.5
WE: Consulting (Computer Hardware)
Followers: 1

Kudos [?]: 80 [0], given: 39

Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]

### Show Tags

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
Math Expert
Joined: 02 Sep 2009
Posts: 35932
Followers: 6860

Kudos [?]: 90101 [2] , given: 10413

Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]

### Show Tags

07 Sep 2014, 11:13
2
KUDOS
Expert's post
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.
_________________
Manager
Status: folding sleeves up
Joined: 26 Apr 2013
Posts: 157
Location: India
Concentration: Finance, Strategy
GMAT 1: 530 Q39 V23
GMAT 2: 560 Q42 V26
GPA: 3.5
WE: Consulting (Computer Hardware)
Followers: 1

Kudos [?]: 80 [0], given: 39

Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]

### Show Tags

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!!
Intern
Joined: 28 Mar 2014
Posts: 4
Schools: Rotman '17
Followers: 0

Kudos [?]: 2 [0], given: 31

Joshua and Jose work at an auto repair center with 4 other [#permalink]

### Show Tags

14 Oct 2014, 09:44
1
This post was
BOOKMARKED
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?
Math Expert
Joined: 02 Sep 2009
Posts: 35932
Followers: 6860

Kudos [?]: 90101 [0], given: 10413

Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]

### Show Tags

14 Oct 2014, 10:07
Expert's post
1
This post was
BOOKMARKED
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.
_________________
Manager
Joined: 30 Mar 2013
Posts: 137
Followers: 0

Kudos [?]: 39 [0], given: 196

Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]

### Show Tags

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) ?
Manager
Status: folding sleeves up
Joined: 26 Apr 2013
Posts: 157
Location: India
Concentration: Finance, Strategy
GMAT 1: 530 Q39 V23
GMAT 2: 560 Q42 V26
GPA: 3.5
WE: Consulting (Computer Hardware)
Followers: 1

Kudos [?]: 80 [0], given: 39

Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]

### Show Tags

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

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
Re: Joshua and Jose work at an auto repair center with 4 other   [#permalink] 15 Oct 2014, 10:49

Go to page    1   2    Next  [ 24 posts ]

Similar topics Replies Last post
Similar
Topics:
1 Jack and Jill work at a hospital with 4 other workers. For an internal 2 02 Jun 2016, 03:33
2 In a card game, a combination of two aces beats all others. If Jose is 1 14 Sep 2015, 21:12
10 Jack and Jill work at a hospital with 4 other workers. For a 6 14 Jun 2013, 10:19
4 Joshua and Jose work at an auto repai 8 14 May 2013, 10:30
1 Work Problem: Repairing a Road 2 19 May 2010, 18:26
Display posts from previous: Sort by