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1. Why trader and Maven? 2. An incomplete deck of cards: if you divide it by 3 you have 2 as remainder; if you divide it by 5 you have 2 as remainder; if you divide it by 4 you have 3 as remainder. How many cards are there in the deck?
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Graduate Trader

Interviewed at Maven Securities

4.3★
Nov 25, 2018

1. Why trader and Maven? 2. An incomplete deck of cards: if you divide it by 3 you have 2 as remainder; if you divide it by 5 you have 2 as remainder; if you divide it by 4 you have 3 as remainder. How many cards are there in the deck?

Asked three arithmetic and probability questions. Honor code of not using a calculator or paper and pen/pencil. Arithmetic: 1. 18/5 2. 76+57 3. 99*99 Probability: 1. A fair coin is flipped three times, what is the probability that it will land on heads at least once.

Intern Assistant Trader

Interviewed at Gray Whale

4.6★
Oct 23, 2012

Asked three arithmetic and probability questions. Honor code of not using a calculator or paper and pen/pencil. Arithmetic: 1. 18/5 2. 76+57 3. 99*99 Probability: 1. A fair coin is flipped three times, what is the probability that it will land on heads at least once.

1) How many times do you have to flip a coin such that the probability of getting 2 heads in a row is at least 1/2? answer: 4 2) You have 1 fair coin and 1 coin with 2 heads. Given that the first flip was a heads what is the probability of getting another heads? Answer: 5/6 3) We are sending 1s and 0s. And there is a 10% chance that the wrong number is sent so we send the number 3 times and the opposing party takes the majority of the 3 tries to be the correct message. What is the probability that we send the wrong message? Answer: 0.028 4) How many sets of A,B,C are there such that they are all non-negative integers and A+B+C<10. Answer: 13C3 = 286. Basically you have 3 dividers and 10 balls at first, but if you add a fourth divider such that A+B+C+D=10, then you've basically got a standard balls and bins problem with 10 balls and 4 bins.
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Trader

Interviewed at Five Rings

3.8★
Oct 7, 2015

1) How many times do you have to flip a coin such that the probability of getting 2 heads in a row is at least 1/2? answer: 4 2) You have 1 fair coin and 1 coin with 2 heads. Given that the first flip was a heads what is the probability of getting another heads? Answer: 5/6 3) We are sending 1s and 0s. And there is a 10% chance that the wrong number is sent so we send the number 3 times and the opposing party takes the majority of the 3 tries to be the correct message. What is the probability that we send the wrong message? Answer: 0.028 4) How many sets of A,B,C are there such that they are all non-negative integers and A+B+C<10. Answer: 13C3 = 286. Basically you have 3 dividers and 10 balls at first, but if you add a fourth divider such that A+B+C+D=10, then you've basically got a standard balls and bins problem with 10 balls and 4 bins.

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