NEW HERE? LEARN MORE
closeclose
Last visit was: 25 Apr 2024, 16:08 It is currently 25 Apr 2024, 16:08
Decision Tracker
My Rewards
New posts
New comers' posts

Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

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

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
aiming700plus
Intern
Intern
Joined: 11 Jun 2006
Posts: 15
Own Kudos [?]: 46 [44]
Given Kudos: 0
Send PM
Joshua and Jose work at an auto repair center with 4 other [#permalink] New post   Updated on: 14 Jul 2014, 02:28
1
Kudos
43
Bookmarks
00:00
A
B
C
D
E

Difficulty:

15% (low)

Question Stats:

77% (01:16) correct 23% (01:18) wrong based on 944 sessions
Hide Show 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
ShowHide Answer
Official Answer
A

Originally posted by aiming700plus on 27 Jun 2006, 22:20.
Last edited by Bunuel on 14 Jul 2014, 02:28, edited 1 time in total.
Edited the question and added the OA.
Most Helpful Reply
User avatar
jaynayak
Director
Director
Joined: 30 Mar 2006
Posts: 896
Own Kudos [?]: 593 [19]
Given Kudos: 0
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post   28 Jun 2006, 00:28
13
Kudos
6
Bookmarks
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
User avatar
L
BrentGMATPrepNow
GMAT Club Legend
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6820
Own Kudos [?]: 29930 [7]
Given Kudos: 799
Location: Canada
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post   13 May 2019, 11:38
2
Kudos
5
Bookmarks
Expert Reply
Top Contributor
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


P(Joshua and Jose both chosen) = P(one of the two men is chosen first AND the other man is chosen second)
= P(one of the two men is chosen first) x P(the other man is chosen second)
= 2/6 x 1/5
= 1/15

Answer: A

Cheers,
Brent
General Discussion
User avatar
trunksli
Intern
Intern
Joined: 06 Apr 2010
Posts: 46
Own Kudos [?]: 35 [6]
Given Kudos: 13
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink] New post   16 Aug 2010, 22:25
5
Kudos
1
Bookmarks
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
User avatar
ezhilkumarank
Senior Manager
Senior Manager
Joined: 24 Jun 2010
Status:Time to step up the tempo
Posts: 273
Own Kudos [?]: 673 [2]
Given Kudos: 50
Location: Milky way
Concentration: International Business, Marketing
Schools:ISB, Tepper - CMU, Chicago Booth, LSB
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink] New post   16 Aug 2010, 18:05
2
Kudos
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.....
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619022 [2]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post   07 Sep 2014, 12:13
2
Kudos
Expert Reply
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.
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619022 [2]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post   14 Oct 2014, 11:07
1
Kudos
1
Bookmarks
Expert Reply
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.
User avatar
mainhoon
Director
Director
Joined: 18 Jul 2010
Status:Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Posts: 538
Own Kudos [?]: 360 [1]
Given Kudos: 15
Concentration: $ Finance $
Schools:Wharton, Sloan, Chicago, Haas
 Q50  V37
GPA: 4.0
WE 1: 8 years in Oil&Gas
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink] New post   16 Aug 2010, 18:45
1
Kudos
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.
avatar
pankajattri
Intern
Intern
Joined: 03 Aug 2010
Posts: 9
Own Kudos [?]: 51 [1]
Given Kudos: 0
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink] New post   17 Aug 2010, 02:23
1
Kudos
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
User avatar
ingoditrust
Senior Manager
Senior Manager
Joined: 14 Mar 2010
Status:Fighting on
Posts: 257
Own Kudos [?]: 58 [1]
Given Kudos: 3
Concentration: Product management/entrepreneurship
Schools:UCLA (R1 interview-WL), UNC(R2--interview-ding) Oxford(R2-Admit), Kelley (R2- Admit $$), McCombs(R2)
GMAT 2: 700  Q48  V38
GPA: 4.0
WE 1: SE - 1
WE 2: Engineer - 3
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink] New post   17 Aug 2010, 16:59
1
Kudos
total ways to select 2 people out of 6 = 6c2 = 15
the number of ways to select J & J = 2c2 = 1

= 1/ 15 (answer)
avatar
nyoebic
Intern
Intern
Joined: 28 Mar 2014
Posts: 4
Own Kudos [?]: 5 [1]
Given Kudos: 31
Schools: Rotman '17
Send PM
Joshua and Jose work at an auto repair center with 4 other [#permalink] New post   14 Oct 2014, 10:44
1
Bookmarks
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?
User avatar
email2vm
Manager
Manager
Joined: 26 Apr 2013
Status:folding sleeves up
Posts: 101
Own Kudos [?]: 702 [0]
Given Kudos: 39
Location: India
Concentration: Finance, Strategy
GMAT 1: 530 Q39 V23
GMAT 2: 560 Q42 V26
GPA: 3.5
WE:Consulting (Computer Hardware)
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post   07 Sep 2014, 12:06
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
User avatar
email2vm
Manager
Manager
Joined: 26 Apr 2013
Status:folding sleeves up
Posts: 101
Own Kudos [?]: 702 [0]
Given Kudos: 39
Location: India
Concentration: Finance, Strategy
GMAT 1: 530 Q39 V23
GMAT 2: 560 Q42 V26
GPA: 3.5
WE:Consulting (Computer Hardware)
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post   07 Sep 2014, 12:36
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!!
User avatar
P
jabhatta2
VP
VP
Joined: 15 Dec 2016
Posts: 1374
Own Kudos [?]: 207 [0]
Given Kudos: 189
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post   15 Sep 2017, 12:27
Hi - why doesnt the following work

(1/6) * (1/5) = 1/30

just wondering why this didnt work ...
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619022 [0]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post   16 Sep 2017, 04:24
Expert Reply
jabhatta@umail.iu.edu wrote:
Hi - why doesnt the following work

(1/6) * (1/5) = 1/30

just wondering why this didnt work ...


What is the logic behind this?

If you say that 1/6 is the probability of choosing Joshua first and 1/5 is the probability of choosing Jose second, then you are missing a case when you are choosing Jose first and Joshua second, so 1/6*1/5 + 1/6*1/5 = 1/15.
User avatar
B
ManoramaS
Intern
Intern
Joined: 07 Sep 2020
Posts: 11
Own Kudos [?]: 2 [0]
Given Kudos: 77
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post   28 Jul 2023, 04:19
Bunuel Please help me why 1-4/6*3/5 is wrong ?
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619022 [0]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post   28 Jul 2023, 04:34
Expert Reply
ManoramaS wrote:
Bunuel Please help me why 1-4/6*3/5 is wrong ?


Hi,

Your doubt is already addressed here: https://gmatclub.com/forum/joshua-and-j ... l#p1407022

Hope it helps.
GMAT Club Bot
gmatclubot
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]
28 Jul 2023, 04:34
Moderators:
Bunuel
Math Expert
92915 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions