Trading Assistant Interview Questions

841

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.

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.

Do you mean Prob(at least 1 observation of 2 consecutive heads appears) ?? Then since the subsets can be: 1. all tails 2. 1 head 1 tail 1 head 1 tail .... 3. at least 2 heads consecutively appear once So Pr(3) = 1- Pr(1) - Pr(2) = 1- (1/2)^N - (1/2)^N = 1 - power(1/2,N-1)

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))

Only solution I could think of, which took quite some time, was to actually start by drawing a tree of possible outcomes. From there, I could see only three different states at each node. Either I am in a state similar to my initial state (when I toss a tail), I can also be in tilt state (I have just flipped heads after flipping a tail) or I am in the done state (I have just flipped heads after having flipped heads before). I consider that someone would only continue flipping if he/she has not flipped two heads in a row. That is, is not in the done state. That is, for any toss level N, it wrong to assume that number of nodes is equal to 2^N. It will be strictly smaller as of level N = 3. Anyway, after 3-4 levels drawn, one should see a pattern, that is, the number of tails, or initial state nodes (is), at a level N is equal to the number of is's I had at level N-1 + the number of is's I had at level N-2. We can thus write: is(N) = is(N-1) + is(N-2). we then realize that at every level N, the number of tilt nodes (t) is equal to the number of initial states from the previous level. That is: t(N) = is(N-1) That makes sense, because on the level N-1, all initial states output one tail (becoming a new initial state at level N) and a heads (becoming a tilt state at level N). the other states of level N-1 are either done (so no continuation) or tilt. The possible outcomes of a tilt state are either done or initial state (if I flip heads again, I am done, so I can't go from tilt to tilt.) One then realize that following the same logic, the number of done (d) nodes for a level N is given by: d(N) = t(N-1) = is(N-2) that is, at every level (as of the second level, or two tosses), I will have a number of done outcomes equal to the number of initial states to levels (or tosses) ago. So the general recursive formula to get the number of done nodes for a certain level really is: d(N) = is(N-2) = is(N-3) + is(N-4) We note, by definition: is(N) = 0, is N<0 is(N) = 1, if N = 0, or 1 is(N) = is(N-1) + is(N-2) otherwise. Furthermore, the probability of being at any node of a level N is simply (1/2)^N. So if I want to find the probability that I observed at least one occurrence of 2 heads in a row. I simply add the number of done nodes from level 1 to N multiplying them by the probability of them occurring at each level. That is: P(hh after N tosses) = sum(from i = 1 to N) (d(i).(1/2)^i). E.g. for N = 2, I get: d(1).1/2 + d(2).(1/4) = s(-1).1/2 + s(0).(1/4) = 0 + 1/4. We know that this is true. you can test it for a larger number of tosses and see it is true. as N goes to infinity, the sum will converge to 1, as I have theoretically no chance of never tossing 2 heads in a row if I throw a coin an infinity number of times. I haven't found a solution that didn't use recursions. There probably is one, I'd be interested in seeing one.

anything below $387.50

Don't be surprised if you have not been employed. The question is "what is the expected return of this game?": E[X]=Sum(P(Xi)*Xi)=1/6*(-1*100+2*100+...)=50$ Esco910, we make a deal when you want, I have always dreamed about to buy a ferrari

2 is a prime number, and the expected value is less than 0. So I wouldn't play this game.

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 :)

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

Adam is right. 7.5 degrees.

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%

The question asked for you to calculate the odds, not the probability.

360 degrees in a clock, divide amongst the hours - you can guess the rest. 1 red marble in one jar and 49 red marbles and 50 white marbles in the second jar. 1/2*1 + 49/99*1/2 = answer.

Overall, the key is to be confident and relaxed. They are just testing how you handle a hostile work environment.