Hacker News new | comments | show | ask | jobs | submit login
Microsoft AI Interview Questions – Acing the AI Interview (medium.com)
107 points by vimarshk 5 months ago | hide | past | web | favorite | 13 comments



To get a feel for how Microsoft is thinking about AI, quite a few episodes of the Microsoft Research podcast are handy too: https://www.microsoft.com/en-us/research/blog/category/podca... .. if you really are going to interview at Microsoft, having an idea of what their researchers are doing might be a huge help.


Thank you for the link. Can I add it to my article quoting you?


Sure, go for it :)


Thanks!


I'm dumb. The probability of rainy Seattle would be (8/27)P(R) where P(R) is the prior probability of rainy weather, no?


Ignoring the P(Rain|Seattle), here's a way to look at this: let's say it's not raining. Then all three must be lying; the probability of that is 1/27. So the probability that it's raining, given that these three are saying it's raining, is 26/27


Either all three must be lying, or all three must be telling the truth, right? So wouldn't you have to discard the probabilities of them disagreeing? There would be a 1/27 chance of all lying, and wouldn't there only be an 8/27 chance of all telling the truth? The other 18/27 would be a mixture of lies and truths. So really, if we're just comparing the odds of those two outcomes to each other, would the probability just be 8/9 that it is raining in Seattle?

Been a while since I've done anything with probabilities, so perhaps I'm way off here.


You CAN'T ignore the extra information like P(Rain|Seattle) though. What if this were the question instead: "Three friends in Seattle told you that Earth exploded and everyone died. Each has a probability of 1/3 of lying. What’s the probability that the Earth exploded and everyone died?"


The correct answer. But the question supposes one can know with accuracy what the percent chance that a given statement is a lie, which imo is pretty ridiculous. Fine abstract mathematical problem, but not very useful


This is my reasoning, as well.


That doesn't work: if twenty of your friends told you it was raining, your method would tell you the probability was (2/3)^20 P(R), which is very small.

Your method was my first thought too - it took a while before I saw how to do it.


The replies to this don't look correct so here's my take:

   P(R) = Probability of rain in Seattle
   P(F) = Probability of three friends saying it's raining in Seattle

   P(F) = P(F|R)*P(R) + P(F|R')*(1-P(R))
   = (8/27)*P(R) + (1/27)*(1-P(R))
   = (7/27)*P(R) + (1/27)

   P(R|F) = (P(F|R)*P(R))/P(F)
   = [(8/27)*P(R)] / [(7/27)*P(R) + (1/27)]
   = 8*P(R) / (7*P(R) + 1)
So, for instance, if the true probability of rain in Seattle is 50%, the probability of it raining in Seattle given our three friends are saying that it is raining in Seattle is 4/4.5 = 89%


Does anyone have a recommended reading list to accompany this?




Applications are open for YC Winter 2019

Guidelines | FAQ | Support | API | Security | Lists | Bookmarklet | Legal | Apply to YC | Contact

Search: