Trading Assistant Interview Questions
556
Trading Assistant interview questions shared by candidates
I have 20% chance to have cavity gene. If I do have the gene, there is 51% chance that I will have at least one cavity over 1 year. If I don’t have the gene, there is 19% chance that I will have at least one cavity over 1 year. Given that I have a cavity in 6 months, what’s the probability that I have at least a cavity over 1 year?
7 Answers↳
You can use the Bayes theorem to calculate the posterior probability of having the cavity gene. P(gene | cavity 1) = P( cavity | gene) * P( gene) / P(cavity) = 2/5 (rounding off all the probabilities) Now, using the posterior probability, you can calculate the revised probability of having a cavity P( cavity 2 | cavity 1) = P( cavity 2 | gene) * P( gene | cavity 1) + P( cavity 2 | no gene) * P( no gene | cavity 1) = 8/25 Less
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I also got 229/700
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First calculate p(G | C in 6 M) = p(C in 6 M| G) p(G) /( p(C in 6 M| G)p(G) + p(C in 6 M | !G) p(!G) ) = 3/7 Now p(C in 1 Y| C in 6 M) = 0.51 * 3/7 + 0.19 * 4/7 = 2.39/7 Less
What is the expected value to you of the following game: You and another person have 3 coins each. You both flip all of your coins and if your three coins show the same number of tails as the other person, you pay them $2, otherwise they pay you $1.
6 Answers↳
Here's the explanation to the above (Since none was given): There are four different number of Tails that can be shown when you flip 3 coins: 0,1,2,3. The Probability of 0 Tails, P(0T) = 1/16, P(3T) = 1/16, P(1T) = 3/16, P(2T)=3/16. Since your flips and your opponent's flips are independent, we know that the Probability that I roll 0 tails and you roll 0 tails = P(0T)*P(0T). Thus, square each of the above probabilities and sum them: you get 1/64+1/64+9/64+9/64=20/64. Thus P(same number of Tails) = 20/64 = 5/16. Now it is asking you for expected value, so plug the above probability into an expected value equation: (5/16)(-2) + (11/16)(1) = 1/16 ie, 5/16 of the time you roll the same number of tails and lose two dollars, 11/16 of the time you don't and you win a dollar, thus your expected value of each play of the game is 1/16 of a dollar. (In the long run if you play many times you should expect to win that much per play) Less
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The above has the correct answer, but the probabilities are 1/8, 1/8, 3/8, 3/8. Then square them....and so on. Less
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Probabilities are: p(0)=1/8, p(1)=3/8, p(2)=3/8, p(3)=1/8.
What the probability of getting 2 consecutive heads in a total of N tosses (I found this one pretty hard and I didn't figure out the right answer at the time.)
6 Answers↳
I think the answer should be this: Since the probability of getting two heads in a row and getting two tails in a row is the same, we only need to figure out the probability of the total. The remaining event is: 1 head 1 tail......due to different order, events should be 2 Therefore the probability=0.5(1-2(1/2)^N)=0.5(1-(1/2)^(N-1)) Less
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sorry for all the typos, the solution should be understandable in spite of that.
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This can be solved by considering the number of sequences of heads/tails that do NOT have consecutive heads. If N coins are tossed, you can have 0 heads in 1 way. You can have 1 head in N ways--choose 1 from any of the N positions in the sequence. You can have 2 heads as long as they are not consecutive. Imagine you have N-2 tails and you need to decide where to place the 2 heads. You can insert them before any of the tails, or after the last tail--so you need to choose 2 out of a possible N-1 positions. Similarly, for any H less than or equal to N/2, you can have H heads by selecting H positions from a possible (N-H)+1 positions. So the number of sequences which do NOT have two consecutive heads can be found by the sum: 1 + nCr(N,1) + nCr(N-1,2) + nCr(N-2,3) + ... Evaluating this starting with N = 2 gives the values 3, 5, 8, 13, 21, 34, ... These are Fibonacci numbers. The number of sequences of length N without 2 consecutive heads is given by F_{N+2}, where F_1 = 1, F_2 = 1, and F_N = F_{N-1} + F_{N-2}. It follows that the probability for obtaining two consecutive heads in N flips of a fair coin is given by 1 - ( F_{N+2}/ 2^N). Note: j-dw has a correct solution. The solution given by Charles ignores the fact that many sequences will have BOTH two consecutive tails AND consecutive heads. He treats these as non-overlapping sets. Delusion ignores all cases such as HTTHTTHTT in which there is more than one T separating the heads. Less
Roll a fair dice, if it's a prime number, you pay me that number times $100, if it's not a prime number, I pay you that number times $100. Consider 1 as a prime number. How much willing to pay to play this game.
5 Answers↳
2 is a prime number, and the expected value is less than 0. So I wouldn't play this game. Less
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$50 is correct, the others are wrong. You would play this infinitely many times because on average, you are printing money. Less
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2 is definitely a prime number... the expectation is less than 0
If you have a deck of cards split into 4 piles and was offered 1:1 odds to draw a face card (J Q K A) from at least one of the piles, would you take the game? Why or why not?
4 Answers↳
If there are 4 piles and you can look at all of them then it's obvious that you're going to take it - all the cards are there. If the question is to take the top card of each of the piles, then it's equivalent to finding the probability of a face card in 4 cards dealt. That is equal to 78.25%. So you would take the bet. (Total = 52 choose 4, Not Wanted = 36 choose 4, P(A face card) = 1-Total/Not Wanted) This is different from the answer given by the previous person because the numerator is wrong: he started at 39 when there are in fact 36 non-face cards (16 face cards - 52-16=36) If, however, the question is to pick one of the piles and then see if it has a face card, it is equivalent to randomly shuffling the deck and seeing if there is at least one face card in the first 13 cards. The probability of not having a face card is 0.363% and thus, the probability of having a face card is 99.637% - this means you should take the bet. So while the question can be interpreted in different ways, you should find it that taking the bet is a good proposition. Another way to look at it is as follows: if we assume sampling with replacement, to make calculations easy, we basically have 4 (or 13, depending on how you view it), chances to pick at least one card with probability 4/13 of the deck. This means that the probability of what is asked is almost equal to 1-(9/13)^4=77.04% (really close to the first case) or almost equal to 1-(9/13)^13=99.16% (also really close to the first number). Less
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I'm no expert but I don't see it that way. You lose the game if you draw non-face cards from all 4 piles. The probability of that is 39/52*38/51*37/50*36/49 = 30%. So you have a 70% chance of winning the game. Less
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Yes, play the game at 1:1 odds. There are 16 face cards and 4 equal piles of 13 cards. By adding the probabilities of drawing a face card from at least 1 pile, you get 16/13. (1/13 + 1/13 + 1/13 + 13/13, or 4/13 + 4/13 + 4/13 + 4/13, and so on)...Doesn't matter how the face cards are arranged in the 4 piles. Thus, you would be willing to play the game at 1:1 odds. Less
What is 29^2?
4 Answers↳
Mental math. You could do it mentally. 29 is close to 30. So the upper bound is 30*30 = 900. You would have to subtract 30 and 29. 900- 59 = 841. Less
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Easiest way I can think of is to mentally calculate (30-1)(30-1) and then mentally foil the answers. You end up getting 900 - 30 - 30 + 1 = 841 which is pretty easy to do in your head. Less
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or, take 29 x 30 to get 870. Take away one 29, get 841. Same idea, but more elegant. You've multipled 29 thirty times, you only need 29 of them, so take one away. Less
How many degrees separate the hands on an analog clock at 3:15?
3 Answers↳
the answer is 7.5. the minute hand will be on the 3, the hour hand will be 25% of the way from 3 to 4. there are 30 degrees in each hour (360/12), so 30/4 = 7.5 Less
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Adam is right. 7.5 degrees.
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I would say it's 5 degrees, though momentarily I would be tempted to say zero. Basically my maths is as follows: 1) Since 15mins also point at 3. We need to find out the small shift in the Hour Hand that was made to represent this 15mins. 2) 360d/12 = 20d (this represent the degress between each hour) 3) 20d/4 = 5 degrees (to represent the 15mins shift). Hope i'm correct. Open to corrections if any :) Less
What's 2/14? what's 1/16? 27x33? if 5x5x5 rubix cube is sitting on a desk, exactly how many cubes have only one side touching the air.
3 Answers↳
1) .1428 2).0625 3)891 4)61
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Wow: you can use a calculator, congrats. Too bad you incorrectly answered the only one that requires any real creativity ... if 5x5x5 rubix cube is sitting on a desk, exactly how many cubes have only one side touching the air = 57 Less
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5x3x3 + 4x3 = 57
How many "1's" are are there between 0 and 1000000?
2 Answers↳
There are 9^6 numbers that don't have any digits with 1 in them. So the answer is 10^6 - 9^6. Less
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The question is unclear. I thought it meant when you write the numbers out, how many ones appear. For example, in the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, there are a total of 4 ones that appear, because 11 has 2 ones. In this case, the answer is: (6, 1) * 10^5 + (6, 2) * 10^4 + (6, 3) * 10^3 + (6, 4) * 10^2 + (6, 5) * 10 + (6, 6) + 1 where (n, k) denotes n choose k. To see why, note that we can form six blanks: A B C D E F -- where each blank can take on values 0-9. This represents a valid number between 0 and 999 999. The total number of ways to have k ones is (6, k) * 10^(6-k). So, sum over this quantity from 1 to 6. Finally, we have to add 1 to the result, since 1 000 000 has one 1, and it can't be represented by A B C D E F. Less
What are odds of getting EXACTLY two (2) heads if you flip a fair coin 10 times?
2 Answers↳
combinatorial problem: let T = total number of flips, n= number of heads you want; then the number of possible ways= T!/[(T-n)!*n!]. thus 10!/(2!*8!)=45 the total number of flips= 2^10 (assuming fair coin, even odds). Thus the odds are 45/2^10~ 4.3% Less
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The question asked for you to calculate the odds, not the probability.