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

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27 Jun 2006, 22:20

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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?

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

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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

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

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16 Aug 2010, 18:05

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

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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]

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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]

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17 Aug 2010, 21:40

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.

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

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13 Jul 2014, 22:57

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.

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

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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

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.
_________________

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

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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

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?

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?

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

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14 Oct 2014, 12:46

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) ?

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

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15 Oct 2014, 11:49

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

final answer =

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

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

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15 Oct 2014, 13:57

email2vm wrote:

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.

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

final answer =

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 [/quote]

Got it thanks. but what is wrong with my logic? if you could help, that'll be great. It'll help me conceptually.

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