Quantitative Interview questions and answers

[koupparis]

I got 1/100 using a binomial tree approach for the first question.

 

[koupparis]

Second Question seems easy: No
0/1
1/2
2/3
2/4
3/5
4/5

Last number should be 4/6 ... in which case you don't have a contradiction, yet. I believe the answer is yes, you have to pass through 75%.

 

[aaronhotchner]

I figured it would be no at first, but I wasn't able to do it.

I'm not sure why you would get p = .01 for the first question. Unless I'm doing something wrong, 50 should be the most common result.

 

[koupparis]

Turns out it's not the case. They all have equal probability. Coded it up too and I'm getting about 0.01.

 

[aaronhotchner]

Haha, you're right, I had a typo in my code. I was effectively using (1-p) instead of p for successes and getting a normal-ish distribution as a result.

 

[euroazn]

Last number should be 4/6 ... in which case you don't have a contradiction, yet. I believe the answer is yes, you have to pass through 75%.

Oops, massive brainfart. At least it got me to see why the answer is yes -
Your free through percentage will be some x/(x+n).

We can see there is an x so that x/(x+n)=75%
since
4x=3(x+n) has a solution where x=3n. Nifty.

 

[koupparis]

Oops, massive brainfart. At least it got me to see why the answer is yes -
Your free through percentage will be some x/(x+n).

We can see there is an x so that x/(x+n)=75%
since
4x=3(x+n) has a solution where x=3n. Nifty.

Of course it has a solution when x = 3n. But this is only sufficient, and not a necessary condition.

 

[euroazn]

Of course it has a solution when x = 3n. But this is only sufficient, and not a necessary condition.

Care to elaborate?

 

[koupparis]

Can you guarantee that at some point x = 3n, for every possible combination? Suppose my current ratio is 54 / (54 +92), how do you know that at some point x = 3n? Can I go from 74.81% to 75.62% without touching 75%?

Turns out this was pretty trivial...

Let x = success, y = failure. We're given that our average is less than 75% at point point and then crosses to above 75%. Let's assume we are at the point when another success will get us at 75% or more. Let's represent this as x/(x+y) < .75. Assume that the next roll is a success and that (x + 1) / (x + y +1) > .75. Then the first inequality gives us x < 3y, and the second gives us x > 3y - 1. So 3y -1 < x < 3y, or -1 < x - 3y < 0.

However, since x and y are both integers this inequality is never satisfied, hence a contradiction, hence our assumption (x + 1) / (x + y + 1) > .75 is false. So (x + 1) / (x + y + 1) = .75.

 

[aaronhotchner]

I'm pretty sure you can guarantee exactly that. Imagine a continuous version of this problem. x = 3n is still a solution, so in a continuous world the answer would be absolutely, yes, you must pass through 75%. The discrete problem must therefore "skip" the point at which the success percentage is exactly 75%, but it will still hit x = 3n, so it will still hit 75%.

 

[koupparis]

I'm pretty sure you can guarantee exactly that. Imagine a continuous version of this problem. x = 3n is still a solution, so in a continuous world the answer would be absolutely, yes, you must pass through 75%. The discrete problem must therefore "skip" the point at which the success percentage is exactly 75%, but it will still hit x = 3n, so it will still hit 75%.

For any \(\epsilon>0\)? I agree intuition says yes, but a proof/solution solidifies it.

 

[euroazn]

I was just thinking that since we know a solution exists, e.g x=3n, then we are good to go simply by the intermediate value theorem.

 

[koupparis]

This function is not continuous so IVT doesn't apply.

 

[GLawler]

You are playing basketball and make your first shot, but miss your second. From then on, the probability that you make your next basket is equal to the proportion of shots you've made so far. What is the probability that you make exactly 50 of your first 100 shots?

-----------
This is equivalent to a well known problem that goes under the name Polya's urn. After 100 shots the number of baskets
that you have made is an integer between 1 and 99. By induction one can show that all of these numbers are equally likely. So the probability is 1/99.

 

[euroazn]

Ah, but that depends on what you consider the function do be!
in general functions of the form x/x+n are continuous (at least when x is positive) if we consider x as reals so we know that the value .75 will be hit for some x. We're not guaranteed that x is an integer until we use the logic that shows x=3n.

 

[koupparis]

In this case x is an integer and hence our function is not continuous.

 

[euroazn]

Again, we don't need to restrict and simply extend naturally to all x real. The function becomes continious and we are guaranteed a solution - but not an integral solution. We use the other logic to guarantee that any solution we get needs to be integral.

 

[koupparis]

Still don't follow your logic. How does x = 3n imply that x or n are integers?

 

[euroazn]

n has to be an integer by definition, since it's the number of throws missed!

 

[koupparis]

x=3n is true regardless of what x and n are, given your assumptions. If n is an integer, then x must be as well. In which case this function is not continuous. You can't assume it's continuous based on the fact that x is real, then turn around and say x is an integer, but the prior conclusion still holds. It doesn't.