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Quant Internship Interview Questions

Here are some more...
  1. What is the chance of drawing two cards from the top of the deck that turns out to be the same number? (Trader)

The first card can be any number. Given the first card, now there are 3 of the same number in the deck of 51 remaining cards. hence the probability is 3/51.
 
Here are some more...
  1. Two ropes that are non-uniform in composition burn completely in 1 hour. You only have a lighter and those two ropes and you have got to tell me when 45 minutes are done. (Trading internship)

Just to avoid any confusions... does each rope take 1 hour to burn completely? Or does each rope take 30 minutes to burn completely?
 
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6. You own property that may or may not have oil underneath. If it does have oil it is worth 100k (20% prob), if it doesn't it is worth 30k. What is the expected value of the land and how much would you sell a contract for this land with a strike of 40k? (S&T internship)

Expected value = [(100k)*0.2 + (30k)*(1-0.2)] = 44k

For the contract with a strike 40k, I would charge 4k + Commissions + Risk premium (as there is only one chance)
 
D

Deleted member 2387

Light both ends on first rope & one end of second rope. First rope will take 30 mins to burn this way. When it is gone, light the other end of second rope, which is half done. The second rope will be gone in the next 15 mins. So this way when both ropes are gone it'll be 45 mins.
 
Here's the probability associated with each outcome of rolling two dice:

Outcome Probability
2 1/36
3 2/36
4 3/36
5 4/36
6 5/36
7 6/36
8 5/36
9 4/36
10 3/36
11 2/36
12 1/36

Hence, expected value = Sum_i (outcome_i*Probability_i) = [Sum_{i=2}^{7} i*(i-1) + Sum_{i=8}^{12} i*(13-i)]/36 = 7.
E (1 die) is 3.5
E (2 dice) = 2 E (1 die) = 7.
 
Here's the probability associated with each outcome of rolling two dice:

Outcome Probability
2 1/36
3 2/36
4 3/36
5 4/36
6 5/36
7 6/36
8 5/36
9 4/36
10 3/36
11 2/36
12 1/36

Hence, expected value = Sum_i (outcome_i*Probability_i) = [Sum_{i=2}^{7} i*(i-1) + Sum_{i=8}^{12} i*(13-i)]/36 = 7.
Or... you could do 2*(1+6)/2 = 7, since all the probabilities for one die are the same... and then it's the expected value of a sum of independent variables :)
 

Joy Pathak

Swaptionz
Here is a very interesting question I got asked by an ED recently. It wasn't an official interview but just a "conversation".

Rank these. The criteria for ranking is completely up to you.

Doctor
Trader
Engineer
Bank Teller
Postman
 
Here is a very interesting question I got asked by an ED recently. It wasn't an official interview but just a "conversation".

Rank these. The criteria for ranking is completely up to you.

Doctor
Trader
Engineer
Bank Teller
Postman

I wish there was "Teacher" as well...

Anyway... here's my ranking:

Doctor
Engineer
Trader
Postman
Bank teller

What was your answer?
 
The first card can be any number. Given the first card, now there are 3 of the same number in the deck of 51 remaining cards. hence the probability is 3/51.
Yeah. My initial thought was to say p(a) = 4/52*3/51, which would be true maybe if we were looking for a specific number.

Here's the probability associated with each outcome of rolling two dice:

Outcome Probability
2 1/36
3 2/36
4 3/36
5 4/36
6 5/36
7 6/36
8 5/36
9 4/36
10 3/36
11 2/36
12 1/36

Hence, expected value = Sum_i (outcome_i*Probability_i) = [Sum_{i=2}^{7} i*(i-1) + Sum_{i=8}^{12} i*(13-i)]/36 = 7.

This should be fairly easy if you're a monopoly player. ;)
 
I'm sure the interviewer is more interested in the criteria, how you come up with it and how you use it to rank these titles. The order is irrelevant.
So what line of reasoning were you using? Initially, I was going to rank them by which is closest to the customer but ran into problems between the bank teller and postman. :confused:
 

Joy Pathak

Swaptionz
I should mention... The ED was ex-salomon. Would that affect the answers? lol

I think the question is very interesting because of the point that Andy mentioned. The criteria you use is what really matters. The criteria also gives away what you consider important in relation to other methods of comparison. So it's a ranking within a ranking if you see what I mean.
 
The first card can be any number. Given the first card, now there are 3 of the same number in the deck of 51 remaining cards. hence the probability is 3/51.
Well how can u assume that the first card u picked is a numbered one it can be a king,queen or jack...So i guess the answer is

40/52 * 3/51 another way to look at this is 4C2 *10 / 52C2
 
Well how can u assume that the first card u picked is a numbered one it can be a king,queen or jack...So i guess the answer is

40/52 * 3/51 another way to look at this is 4C2 *10 / 52C2

OK... I had assumed the J, Q, K, A to be in the same category as numbers. You are definitely correct if I have to stick to 1-10.
 
Well how can u assume that the first card u picked is a numbered one it can be a king,queen or jack...So i guess the answer is

40/52 * 3/51 another way to look at this is 4C2 *10 / 52C2
I made a mistake..Numbered cards are from 2-10 so it should be 36/52 * 3/51 another way to look at this is 4C2 *9 / 52C2..
 
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