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

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

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22 Jul 2007, 20:36
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Joshua and Jose work at an auto repair center with 4 other workers. For a survey an health care insurance 2 of the 6 workers were randomly chosen for the interviews. What is the probability that Joshua and Jose would be chosen together?
1/15, 1/12, 1/9, 1/6, 1/3

Here is what I thought

Number of ways to choose 4 workers out of 6- 6C4 = 15

Considering Joshua and Jose were chosen, we need 2 more workers from remaining 4. So, total favorable cases are 4C2 = 6

Hence the probability is 6/15.. which is not in the answer choice..
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22 Jul 2007, 21:04
I though the answer was 1/15.

For first person the prob is 2/6 and and for the second person the prob is 1/5 assuming 1 of them already made it.

so 2/6*1/5 = 1/15.

I am not sure though.
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22 Jul 2007, 21:11
# of ways to pick 2 people from a group of 6 = 6C2 = 15
There is only one way to pick Jose and Joshua. So P = 1/15
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23 Jul 2007, 06:43
ywilfred wrote:
# of ways to pick 2 people from a group of 6 = 6C2 = 15
There is only one way to pick Jose and Joshua. So P = 1/15

I get it thanks!!

argh - I need to read questions carefully!!!
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23 Jul 2007, 06:43
You guys are right.. I was calculating it for 4 out of 6 workers when the question asked 2 out of 6 workers!
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23 Jul 2007, 08:24
Multiply probabilities of events
first event: Jose/Joshua = 1/6
second event: Jose/Joshua, depending on who's left = 2/5

(1/6) * 2/5) = 2/30 = 1/15...
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23 Jul 2007, 21:33
1/15

P josh or Jose are selected in the first position = 2/6 = 1/3
with either one being selected in first position, probability that second position is the remaining is 1/5
P = 1/3*1/5 = 1/15

I am assuming the position does not matter.
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