Simple probability question giving me angst!

Factual Questions
1

Uggh. It’s been a long time since probability class…

What’s the appropriate formula to use to answer this question:

You are a member of a review team of seven department officers. The other six include two administrative officers, two economics officers, and two political officers. If you were to travel with two of them, selected at random, what is the probability that you would be traveling with at least one economics officer?

If I write out all the combinations of two people I arrive at 15 possible combinations with 9 favorable outcomes, so that yields 60%. Which I think is the right answer, BUT I keep wanting to do something along the lines of: Chances that position number 1 is filled with economics officer = 2/6. Chances that position number two will be filled with economics offices (if position one wasn’t filled with an economics officer) is 2/5 because a non-economics officer is now occupying position one. So total probability of at least one being an economics officer is 2/6 + 2/5 or 73%.

Can you lend a brother a hand?

FYI… this was a sample question from a foreign service exam.

2

http://boards.straightdope.com/sdmb/showthread.php?s=&threadid=170830

It’s 60%.

3

I will never again doubt the ability of this board to HAVE ALREADY ANSWERED THE EXACT FREAKIN’ QUESTION that I have. And less than 4 days ago at that!

Thanks for the link Achernar… and thanks to vasyachkin and Valgard from the other thread for the explanantion. I knew I needed to use that 2/5 somewhere!

total = 2/6 + 4/6 * 2/5 = 0.6

4

What you’re missing in your calculation is that you have to discount the possibility of both positions being economics officers. In essence by just adding the probabilities, you’re counting twice all the combination where both companions are economics dudes. So:

Probability of 1st position being EO: 2/6
Probability of 2nd position being EO IF 1st isn’t: 2/5*4/6

Adding those together yields 0.6

Alternately,

Probability of 1st being EO: 2/6

  • probability of 2nd being EO: 2/5
  • probability that both are: 2/6*2/5

Which still adds up to 0.6

5

Of course, there’s the easy way: what’s the probability of having no EOs? 4/6 * 3/5 = 0.40, so the probability of having at least one is 1-0.4=0.6 :slight_smile:

6

Ah, don’t beat yourself up; it was over a month ago. :wink: Anyway, I searched for economics officer to find it; I don’t know how you would have known to do that.