Facebook Brainteasers

1) What is your logic behind 0.5? I would think it would be less than 50% that all 3 random drops would be on the same side of the circle.

1) I'd say 12.5% (5.*.5*.5)
2)The second is a bit harder
All the combos are:
6-3-2 (sum=11)
9-2-2 (sum=13)
4-3-3 (sum=10)
9-4-1 (sum=14)
12-3-1 (sum=16)
18-2-1 (sum=21)
6-6-1 (sum=13)

Since that isn't enough info (an the guy obviously knows the # of beers drunken therefore since 2 add up to 13 (6, 6, 1 and 9, 2, 2) it can only be one of those.
Since there is an oldest the first is not possible so the ages are 9, 2, and 2
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So what do you do?
-I work for an investment banking firm.
Oh okay; you are like my brother, he works for Edward Jones.
-No, a college degree is required in my profession

For the circle, it's more than 0.5.

First point can be anywhere. Second point can be anywhere. For the third point:

best case: the first and second points are either 180 or 0 degrees apart, in which case the third point can be anywhere, so probability is 1.

Worst case: the points are 90 degrees apart, in which case the third point can be on 3/4 of the circle so probability is 3/4.

For all degrees between 0-180, it will vary between 3/4 and 1. Don't have the time to figure it out, I am sure there is some formula for it. Graph would look cos graph, with high and low points at 1 and .75.

edit: shit beaten =(

2 doesn't seem too difficult, solution below (i wish there were a spoiler tag or something that i could black things out

2) Two friends are chatting:
- Peter, how old are your children?
- Well Thomas, there are three of them and the product of their ages is 36.
1, (1,36)(2,18)(3,12)(4,9)(6,6)
2, (1,18)(2,9)(3,6)
3, (1,12)(2,6)(3,4)
4, (1,9)((3,3)
6, (1,6)(2,3)
12,(1,3)
18,(1,2)
36,(1,1)

so that leaves us with 8 distinct sets
(1,1,36)(1,2,18)(1,3,12)(1,4,9)(1,6,6)(2,2,9)(2,3,6)(3,3,4)

- That is not enough ...
- The sum of their ages is exactly the number of beers we have drunk today.
sums are
38, 21, 16, 14, 13, 13, 15, 10
- That is still not enough.
so either (1,6,6) or (2,2,9)
- OK, my oldest child plays guitar.
- Ah, now I understand.
How old were each of Peter's children?
(2,2,9)

now as for #1...

6-6-1 is also a valid response. one of a set of twins is still older.

for the circle question. two of the three will obviously be on the same half...so figured the third is just 50/50.

d/dx of .125cos(2x)+.875 over 0 to pi

Not bad.

like it ideating, i'd agree with your 75%

Shit no, it goes from 0.5 to 1 not 0.75 to 1. Eg. if the two points are 179.9 degree apart it would be 0.5

Given first 2 points being d degrees apart,
p(same half|d) = (360 - d)/360
p(first 2 points being d degrees apart) = 1 / 180 ( d goes from 0 to 180)
Integrating p(same half|d).p(first 2 are d apart) over the interval of d from 0 to 180, I get 75%

How do you guys know they drank 13 beers? Where does it say that in the problem set?

Because the fact that even knowing how many beers they had was not enough to tell the answer. So it had to be the sets with common sums.

Jimbo, by the facebook site do you mean facebook.com? Or your firm's facebook (not sure how common the terminology is firm to firm)?

2 points can at max be the opposite ends of the semi circle.
then the third point is always going to complete the semicircle

It all depends on whether you get to draw the imaginary diameter to cut the circle in half before or after you pick your points.

If the halves pre-determined, the probability will definitely not be one. But I have no idea how to find that probability.

If you get to seperate the halves after you pick your points, probability will definitely be one.

So it's all about how you set your assumptions

i meant facebook.com ideating

I think this has to be far more specific. Is this a circle as in a single, continuous line drawn and the points are somewhere ON that line, or is a circle in the sense of a pie (or something similar) where the dots can be inside the circle? Then it depends on if the diameter is predetermined or not.

If the points are on the line of the circle, it would either be 100% or 0% depending on whether you would consider a point touching the other half of the circle as falling on, or off, the other semi-circle.

Interviewer: You choose three points on a circle randomly, what's the probability that they are all on the same half circle?

Candidate: When you say it's a circle, what exactly do you mean by that?

Interviewer: ...

I was simplifying it a bit too much.
Simple counter example: 3 points 120 degrees apart.

I guess one of you calculus guys might be right.

aren't you guys WAAAAY over complicating this?

Assuming points cannot be on the dividing line and all points must be inside the confines of the circle, Points 1 and 2 are automatically on the same half.

Since it can only have 2 halfs, theres a 1 in 2 chance the 3rd point will be in the same half as points 1 and 2. 50% chance all 3 points will be in the same half.

If the division line is already drawn, then the first part of the solution is not applicable(Points 1 and 2 may not be in the same half). In this case, theres a 1 in 2 chance of each point ending up on side A. .5 * .5 * .5 = .125 = 12.5%

On a side note, I think all these follow-up questions aren't a good idea on an interview. How about just stating your own assumptions, that way, you can choose the assumptions which makes the solution easiest for you to solve.

Read my post above as to why it is not 50%.

Candidate: Uh, so did I get the job?