Post: #1

Hey everyone,

Just got this email from a close friend of mine. Just wanted to see if I could take advantage of the combined brain power here on the forum to help him out. Cheers!

Quote:Hi guys,

I have an interview coming up which will ask me a lot of statistical/probability aptitude questions.

However, unfortunately the ones I could find on the internet are not difficult enough to help me properly prepare. I know there are documents out there which have loads of these to enable people to practice for such interviews. And, i was hoping that someone in my highly skilled circle of friends would have such a document, or could tell me where I could get it.

The questions will all be probability questions designed to see how clever someone is, and are of the type:

"If there is a 50% probability of stars in one night, what is the probability of stars in either of two nights?"

I would greatly appreciate your help on this if you have access to such a document. If not, I hope all is going well with you.

Cardguy

Post: #2

If I'm not misreading the question.

75%

Probably of not having a starry night. Order does not matter. 50% not starry x 50% not starry= 25% Therefore 75% chance of at least one or more of the nights being starry.

Also it can be thought of as a permutation.
S NS
NS S
S S
NS NS

3/4 don't.

If it means Starry given a previous condition P(Night Sky|Night Sky-1), it's still 50%

Post: #3

Whenever you're asked to find the probability of something occurring at least once through a set of events with defined probability, the easiest way to do it is to calculate the probability that it never happens and subtract that from 1.

The probability it never happens is the product of all the individual probability events that it doesn't occur. Here, there's a 50% chance any one night is starry, which means there's a 50% chance any one night is not starry. Thus, the probability that both nights are not starry is 0.5 * 0.5 = 0.25. This means the probability that at least one night is starry is 1-0.25 = 0.75.

Likewise, the probability of at least one night in three being starry would be 1- (0.5*0.5*0.5) = 1-0.125 = 0.875.

If he needs to know the probability of exactly X number of nights being starry out of N total, he needs to use the binomial distribution.

Post: #4

If he needs to know legit probability, then he'll probably need a book.

Post: #5

The book Fifty Challenging Problems in Probability with Solutions by Frederick Mosteller might be what he's looking for. I have a .djvu copy if you want it.

Post: #6

Might want to brush up on set theory in the meantime so you are completely sure what the mathematical definitions for "either", "or", "and", and similar words are.

Post: #7

Has he checked MIT OCW?

He might find good problems if he looks through the course materials they post.

Post: #8

(06-06-2013 12:32 PM)Soma Wrote:  The book Fifty Challenging Problems in Probability with Solutions by Frederick Mosteller might be what he's looking for. I have a .djvu copy if you want it.

Soma,

Could I please get a copy? I have a possible interview coming up which will require knowledge of probabilities. Thanks.


Regards,
SK

Post: #9

(06-06-2013 12:32 PM)Soma Wrote:  The book Fifty Challenging Problems in Probability with Solutions by Frederick Mosteller might be what he's looking for. I have a .djvu copy if you want it.

Cheers! Just PM'ed you.

Post: #10

Heard on the Street: Quantitative Questions from Wall Street Job Interviews

Post: #11

(06-06-2013 02:18 PM)Divorco Wrote:  Heard on the Street: Quantitative Questions from Wall Street Job Interviews

I also have a .djvu copy of this.

Post: #12

(06-06-2013 11:16 AM)lurker Wrote:  Whenever you're asked to find the probability of something occurring at least once through a set of events with defined probability, the easiest way to do it is to calculate the probability that it never happens and subtract that from 1.

The probability it never happens is the product of all the individual probability events that it doesn't occur. Here, there's a 50% chance any one night is starry, which means there's a 50% chance any one night is not starry. Thus, the probability that both nights are not starry is 0.5 * 0.5 = 0.25. This means the probability that at least one night is starry is 1-0.25 = 0.75.

Likewise, the probability of at least one night in three being starry would be 1- (0.5*0.5*0.5) = 1-0.125 = 0.875.

If he needs to know the probability of exactly X number of nights being starry out of N total, he needs to use the binomial distribution.

This. The other guiding rule here is that for 2 INDEPENDENT events, in order to get the probability that they both happen, you multiply the probabilities together.

Not to confuse the issue, but where this often breaks down is in the real world where it's much harder to measure the degree of dependence (i.e. wind currents on cloud cover, condensation, light pollution, etc.)

Which also spills into an oversimplified understanding of risks and probability in society (see the writings of Benoit Mandelbrot and Nassim Taleb).