# Statistics Question **URL:** \ **Category:** Factual Questions\ **Created:** [March 19, 2003, 3:52am UTC](https://boards.straightdope.com/t/statistics-question/162455 "2003-03-19T03:52:18Z")\ **Posts on this page:** 13\ **Page:** 1
**Author:** ![rodent](https://avatars.discourse-cdn.com/v4/letter/r/4bbf92/32.png) [@rodent](https://boards.straightdope.com/u/rodent)\ **Post date:** [March 19, 2003, 3:52am UTC](https://boards.straightdope.com/t/statistics-question/162455/1 "2003-03-19T03:52:18Z")
I saw the following question on a multiple choice test. My answer is different than any of the choices. Can anyone help me determine which answer, if any, is correct?: 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, to your next site, what is the probability that you would be traveling with at least one economics officer. A. .27 B. .40 C. .50 D. .60 (I think it is 5/9, or .56) ---
**Author:** ![jk1245](https://avatars.discourse-cdn.com/v4/letter/j/49beb7/32.png) [@jk1245](https://boards.straightdope.com/u/jk1245)\ **Post date:** [March 19, 2003, 3:59am UTC](https://boards.straightdope.com/t/statistics-question/162455/2 "2003-03-19T03:59:04Z")
Are they rounding the .56 to .6? I get 5/9 also. ---
**Author:** ![Rabid\_Squirrel](https://avatars.discourse-cdn.com/v4/letter/r/e0b2c6/32.png) [@Rabid\_Squirrel](https://boards.straightdope.com/u/Rabid_Squirrel)\ **Post date:** [March 19, 2003, 4:02am UTC](https://boards.straightdope.com/t/statistics-question/162455/3 "2003-03-19T04:02:52Z")
OK, chance of first person is econ, not the second = 2/7 x 4/6 = 8/56 Chance of second person (but not first) is in econ. = 5/7 x 2/6 = 10/42 Adding the two probabilities gives 8/21 or 38% Answer: B Or cynic’s answer, certain since they’re always the worst travellers. Cripes, haven’t done a question like that for 10 years or so… ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [March 19, 2003, 4:02am UTC](https://boards.straightdope.com/t/statistics-question/162455/4 "2003-03-19T04:02:59Z")
The possible combinations are: AA AE AP EA EE EP PA PE PP Since 5 of those 9 have at least one Economics Officer, it looks like you’re right to me. ---
**Author:** ![Goomba](https://avatars.discourse-cdn.com/v4/letter/g/65b543/32.png) [@Goomba](https://boards.straightdope.com/u/Goomba)\ **Post date:** [March 19, 2003, 4:12am UTC](https://boards.straightdope.com/t/statistics-question/162455/5 "2003-03-19T04:12:27Z")
Hmm. Haven’t taken math in a while but here’s my go at it. First choose an officer at random. Notice there is no difference between an administrative or political. Case A: (4 subjects) 1st officer is not economics. The second officer will be an economics officer 2/5 of the time since there are 2 economics officers left out of 5. ``` Since there are four of these, there are 8 out of 20 possibilities where objectives are met. ``` Case B: (2 subjects) 1st officer is economics. The second officer doesn’t matter. No matter which of the 5 is randomly picked, the objectives are met. ``` There are two of these. 10 out of 10 possibilities here. ``` Result: 18/30 reach objective=.6 (letter D) Goomba ---
**Author:** ![jk1245](https://avatars.discourse-cdn.com/v4/letter/j/49beb7/32.png) [@jk1245](https://boards.straightdope.com/u/jk1245)\ **Post date:** [March 19, 2003, 4:14am UTC](https://boards.straightdope.com/t/statistics-question/162455/6 "2003-03-19T04:14:34Z")
Aha, Rabid Squirrel showed me the way. 1st person an econ, not the second 2/6 \* 4/5 = 8/30 2nd, not the first: 4/6\* 2/5 = 8/30 Both 2/6 \* 1/5 = 2/30 Total = 18/30 or 0.6 ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [March 19, 2003, 4:30am UTC](https://boards.straightdope.com/t/statistics-question/162455/7 "2003-03-19T04:30:38Z")
I think the easiest way to do this is like this: Total number of (unordered) pairs of people, out of 6, is 6C2 = 15 Some of these pairs will feature two non-economists. How many? Since there are four non-economists, it’s 4C2 = 6. Thus of the 15 possibilities, 6 will have zero economists, so 9 must have one or two. 9/15 = 0.6. The problem with your analysis, if it’s the same as **Q.E.D.**'s, is that not all nine combinations are equally likely. For instance, AE is twice as likely as AA. ---
**Author:** ![vasyachkin](https://avatars.discourse-cdn.com/v4/letter/v/c5a1d2/32.png) [@vasyachkin](https://boards.straightdope.com/u/vasyachkin)\ **Post date:** [March 19, 2003, 5:55am UTC](https://boards.straightdope.com/t/statistics-question/162455/8 "2003-03-19T05:55:11Z")
probability that first one is econ = 2/6 probability that he isn’t is 4/6 prob that he isn’t but second one is = 4/6 \* 2/5 total = 2/6 + 4/6 \* 2/5 = 0.6 EXACTLY WITHOUT ROUNDING OFF. i think jk1245 did the same thing only he did extra work, i just combined some steps. ---
**Author:** ![Valgard](https://avatars.discourse-cdn.com/v4/letter/v/7feea3/32.png) [@Valgard](https://boards.straightdope.com/u/Valgard)\ **Post date:** [March 19, 2003, 5:59am UTC](https://boards.straightdope.com/t/statistics-question/162455/9 "2003-03-19T05:59:14Z")
Or do it this way: P(at least one economist) = 1-P(no economists) First random pick: 4/6 chance no economist Second random pick: 3/5 chance no economist In this case the events are SI so the probability of no economists in two picks is just the product of the two probabilities. (4/6)\*(3/5) = 12/30 = 0.4 P(at least one economist) = 1-0.4 = 0.6 (D) ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [March 19, 2003, 6:01am UTC](https://boards.straightdope.com/t/statistics-question/162455/10 "2003-03-19T06:01:08Z")
I see I made an obvious error. I had assumed the chances of each case were the same, but reading the later posts, I see that is clearly not the case. ---
**Author:** ![tullius](https://avatars.discourse-cdn.com/v4/letter/t/977dab/32.png) [@tullius](https://boards.straightdope.com/u/tullius)\ **Post date:** [March 19, 2003, 2:04pm UTC](https://boards.straightdope.com/t/statistics-question/162455/11 "2003-03-19T14:04:46Z")
Good luck on the foreign service exam, **rodent**. Any idea whether their questions on statistics cover things like standard deviations? ---
**Author:** ![Scuba\_Ben](https://avatars.discourse-cdn.com/v4/letter/s/278dde/32.png) [@Scuba\_Ben](https://boards.straightdope.com/u/Scuba_Ben)\ **Post date:** [March 19, 2003, 2:06pm UTC](https://boards.straightdope.com/t/statistics-question/162455/12 "2003-03-19T14:06:50Z")
**Valgard** has the best answer. When solving probability problems such as the OP, it may be easier to compute the probability of the given event NOT happening. ---
**Author:** ![rodent](https://avatars.discourse-cdn.com/v4/letter/r/4bbf92/32.png) [@rodent](https://boards.straightdope.com/u/rodent)\ **Post date:** [March 19, 2003, 3:35pm UTC](https://boards.straightdope.com/t/statistics-question/162455/13 "2003-03-19T15:35:16Z")
Thank tullius for the encouragement. Thanks to everyone else for the explanations. I thought I knew statistics pretty well. But this one really threw me.