Last visit was: 14 Jul 2024, 22:43 It is currently 14 Jul 2024, 22:43
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.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
User avatar
Intern
Intern
Joined: 11 Jun 2006
Posts: 15
Own Kudos [?]: 47 [45]
Given Kudos: 0
Send PM
Most Helpful Reply
User avatar
Director
Director
Joined: 30 Mar 2006
Posts: 894
Own Kudos [?]: 604 [19]
Given Kudos: 0
Send PM
GMAT Club Legend
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6805
Own Kudos [?]: 30799 [7]
Given Kudos: 799
Location: Canada
Send PM
General Discussion
User avatar
Senior Manager
Senior Manager
Joined: 24 Jun 2010
Status:Time to step up the tempo
Posts: 273
Own Kudos [?]: 678 [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]
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
Director
Director
Joined: 18 Jul 2010
Status:Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Posts: 536
Own Kudos [?]: 362 [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]
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.
User avatar
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]
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
avatar
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]
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
Senior Manager
Senior Manager
Joined: 14 Mar 2010
Status:Fighting on
Posts: 255
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]
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)
User avatar
Manager
Manager
Joined: 26 Apr 2013
Status:folding sleeves up
Posts: 101
Own Kudos [?]: 712 [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]
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: 94342
Own Kudos [?]: 640866 [2]
Given Kudos: 85011
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]
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
Manager
Manager
Joined: 26 Apr 2013
Status:folding sleeves up
Posts: 101
Own Kudos [?]: 712 [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]
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!!
avatar
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]
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?
Math Expert
Joined: 02 Sep 2009
Posts: 94342
Own Kudos [?]: 640866 [2]
Given Kudos: 85011
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]
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.
VP
VP
Joined: 15 Dec 2016
Posts: 1352
Own Kudos [?]: 220 [0]
Given Kudos: 188
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]
Hi - why doesnt the following work

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

just wondering why this didnt work ...
Math Expert
Joined: 02 Sep 2009
Posts: 94342
Own Kudos [?]: 640866 [0]
Given Kudos: 85011
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]
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.
Intern
Intern
Joined: 07 Sep 2020
Posts: 10
Own Kudos [?]: 2 [0]
Given Kudos: 79
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]
Bunuel Please help me why 1-4/6*3/5 is wrong ?
Math Expert
Joined: 02 Sep 2009
Posts: 94342
Own Kudos [?]: 640866 [0]
Given Kudos: 85011
Send PM
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]
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
Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]
Moderator:
Math Expert
94342 posts