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

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

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