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

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aiming700plus
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Joshua and Jose work at an auto repair center with 4 other [#permalink] New post  Updated on: 14 Jul 2014, 02:28
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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
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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.
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jaynayak
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post  28 Jun 2006, 00:28
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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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BrentGMATPrepNow
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post  13 May 2019, 11:38
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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
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Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink] New post  16 Aug 2010, 18:05
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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.....
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Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink] New post  16 Aug 2010, 18:45
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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.
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Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink] New post  16 Aug 2010, 22:25
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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
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Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink] New post  17 Aug 2010, 02:23
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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
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Re: Joshua and Jose work at an auto repair center with 4 other workers. [#permalink] New post  17 Aug 2010, 16:59
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total ways to select 2 people out of 6 = 6c2 = 15
the number of ways to select J & J = 2c2 = 1

= 1/ 15 (answer)
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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
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Bunuel
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post  07 Sep 2014, 12:13
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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, 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!!
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Joshua and Jose work at an auto repair center with 4 other [#permalink] New post  14 Oct 2014, 10:44
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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, 11:07
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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  15 Sep 2017, 12:27
Hi - why doesnt the following work

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

just wondering why this didnt work ...
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post  16 Sep 2017, 04:24
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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.
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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 ?
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink] New post  28 Jul 2023, 04:34
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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.
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Re: Joshua and Jose work at an auto repair center with 4 other [#permalink]
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