IMC Trading quant interview problem | Frog Return

[@Quant_Prof]

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[@piyushbhujbal3075]

I am still having problems with payment.

[@purple_rabbit19]

I am not preparing to become a quant trader, but I find these problems so fascinating, good work!

[@stefano1774]

*** spoiler for the second problem *** we can fix the two starting point A and B on the x axis, and integrate over the probability that A will meet the trail of B, given the angle taken by B. first of all, we only have to check for angles of B between 0 and 180, because it is obviously symmetric, and the probability distribution is uniform over this "space" of angles, and it's 1/180. secondly, we can easily see that for each angle theta taken by B, the probability that A will meet the trail of B is theta/360 (because A can take every angle between 0 and 360 with equal probability, and it is going to intersect with B for every angle between 0 and theta itself, where the two trajectories are parallel). so we have the integral: integral from 0 to 180 of (1/180)*(theta/360) d theta this is equal to 1/4. I think it's correct and probably there are easier way to get the result, but it was fun to solve anyway!

[@vexrav]

* spoilers * The planes will be traveling parallel with measure zero, so assume that the planes lines intersect, for the rays to intersect both rays have to be pointing towards the point of intersection and 1/2 * 1/2 = 1/4

[@gustavomeza9137]

Solution for second problem. 1. The probability of the "lines" not crossing is zero. 2. Given two "lines" that cross, the probabilty of each "ray" to face the direction of the intersection is 1/2. Hence 1/4

[@maxie7806]

why is the probability 0? what if they are parallel? or what if they start "facing away" from each other? as these are not infinite lines, they have a starting point?

[@anthonyhu6705]

@@maxie7806 1. There’s 0 probability that out of all the possible angles, they are exactly parallel - it’s like the chance of picking 0.5 out of all the real numbers between 0 and 1. 2. Gusta addresses the jets making half lines rather than lines in the second part of his answer

[@maxie7806]

@@anthonyhu6705 thanks for replying. I understand Gusta's 2nd point, but am still confused on the first. also if p(of the lines not crossing) = 0, doesn't that make p(lines cross) = 1??? sorry for the hassle, I understand why the answer is 1/4 but not point 1

[@yeah5037]

Solution for second Problem: Let the A,B be the points at which the planes start. Fix A as origin and the +ve x axis along its trail. Hence we have an x-y plane. Now take any point on +ve y axis. If it's angle is between [-90,0) the trails will intersect. So for these points the probability is 1/4 . Now if we go a bit ahead in first quadrant the required angle will be [-90-theta,0) and there will be a symmetric point in second quad with required angle [-90+theta,0) . If you make such pairs the req prob is again 1/4. Similar symmetry exists in third and fourth quadrant as well. Hence the ans is 1/4

[@aadfg0]

WLOG points on x-axis. Probability 1/2 both rays face same of up/down, WLOG up. Probability 1/2 the left point has smaller angle than the right point, hence answer is 1/4.

[@estring123]

@quant_prof 2nd question is same as asking, give a straight line in an infinite plane, and 2 random angles a1, a2 are selected for its 2 end points, wats the probability of the 2 lines coming out of the 2 end points will form a triangle the sample sample space is [0, 2pi) X [0, 2pi) under normalized lebesgue measure, i.e. divided by 1/4pi^2. the region that allows a triangle to form is 0

[@Quant_Prof]

can you please mail us at [e-mail]?

[@maxie7806]

this looks like an interesting solution which I don't fully understand, do you mind explaining what is meant by the last part? (after "the sample sample...")

[@lovyajain4032]

(180- (theta))/360 where theta has to be avaeraged from 0 to 180 yielding 1/4

[@arpanakumari1875]

Awesome

[@eobardthawne19]

my soln= we must divide the jumps into 2 categories : Left and Right the sum of numbers in these grps must be equal for it to return to origin and now , the sum of these 2 grps must be total distance frog jumps which is just sum from 1 to 50 which is 1275 which is odd and hence cant be split into 2 equal grps thus prob is 0