Probability from GMATPREP software

[guest612]

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.

[tmmyc]

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.

[guest612]

thanks, tmmyc. that really helped!

[StaceyKoprince]

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.

[ivanushk]

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.

Is there a way to solve this problem using 1 - P(NOT)? I know how to do it using the other methods, but the last one does not give me a peace of mind. Please let me know.

The way I think is P(NOT Josh & Jose) = 4/6*3/5=2/5 but then I am not sure what I got, could somebody help me with the steps I need to take to get to 14/15? Thanks in advance.

I understand this method is not as efficient, just want to make sure I got it.

[RonPurewal]

Is there a way to solve this problem using 1 - P(NOT)? I know how to do it using the other methods, but the last one does not give me a peace of mind. Please let me know.

The way I think is P(NOT Josh & Jose) = 4/6*3/5=2/5

That's the probability that neither of the two is selected. "Neither" is not the same as "not both".

You'd also have to find the probability that exactly one of them is selected, and then add that to the 2/5.

[ivanushk]

Is there a way to solve this problem using 1 - P(NOT)? I know how to do it using the other methods, but the last one does not give me a peace of mind. Please let me know.

The way I think is P(NOT Josh & Jose) = 4/6*3/5=2/5

That's the probability that neither of the two is selected. "Neither" is not the same as "not both".

You'd also have to find the probability that exactly one of them is selected, and then add that to the 2/5.

Thanks a lot Ron, this makes total sense now.

On another note, I would like to know a place to ask a follow-up question regarding one of your sessions of Thursdays with Ron (Absolute Values workshop - February 18, 2010), where do I do that please? Or email is the only option? Thanks.

[RonPurewal]

If you know the source of the problem from the video, then post in the corresponding folder.
If you don't, post in the general folder, and cite the date of the video as your source. (The same sources that are banned here are also banned from the videos"”"”and from any other source that we offer as a free service"”"”so the problems should be ok to post here.)