Joshua and Jose work at an auto repair center with four other workers. For a survey on a healthcare insurance, 2 of the 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

The correct answer is A. 1/15.

I tried finding the possibility of Joshua and Jose NOT Being chosen and still couldn't get to it. I did this by 1-(4/6). Can you please help? Thank you.

To get to 1/15, this is how I would think about it.

Probability is basically the number of combinations you want, divided by the total possible number of combinations.

There are 6 total people: Joshua, Jose, and 4 others.
The survey is selecting 2 people out of these 6.

How many possible 2-person combinations are there? We can use Combinations:
6 choose 2
-> 6! / (2! * 4!)
--> (6*5)/(2*1)
--->15 possible 2-person combinations.

Now how many 2-person combinations do we want? Only 1, namely the combination that has Joshua and Jose.

Bringing this all together, we want 1 2-person combination (the Joshua-Jose combination), and there are 15 possible 2-person combinations. The probability of getting the Joshua-Jose combination is then 1/15.
Hence, when the question asks what the probability that Joshua and Jose will both be chosen is, the answer is 1/15.

thanks, tmmyc. that really helped!

Thanks tmmyc. Also, guest612, remember that if you are going to calculate the reverse, you'd have to find the number of ways that Joshua is chosen but Jose is not, the number of ways that Jose is chosen but Joshua is not, AND the number of ways that neither one is chosen. Those comprise all of the ways in which we do not choose BOTH Joshua and Jose. That's a lot more work (because you have to calculate more scenarios!) so it's generally not worth it to do this problem in that way.