Hard | Probability |

N points are chosen at random on the cicumference of a circle. A convex n-gon (n sided polygon) is drawn by joining these n points. What is the probability that the center of circle lies inside the region of n-gon?

Solution

Source: a nice senior

Enable Like and Comment Hard | Probability |

On pressing a button, a random number is generated uniformly between 0 & 1. You keep on generating these numbers until their sum exceeds 1. What is the probability that you need to press the button more than n times? What is the expected number of times you need to press the button?

Hint

Probability by conditioning

Answer

Probability: 1/n! ; Expected times: e

Solution

Probability of more than n throws is equivalent to saying X1+..+Xn <=1, this is the volume of n dimensional symplex from origin, its volume is 1/n! and can be proved by induction. This can be used to calculate expected number of button presses, sum( 1/n! * n ) = sum( 1/(n-1)!) = e^1

Otherwise, set up a (lag)differential equation for f(x), the expected number of draws needed for the sum to exceed x. For x=0,f(x)=1. For x>0, suppose a draw gave number t, then f(x)= 1 + integral (t=0 to x)f(x-t)dt, which is same as f(x) = 1 + integral(t=0 to x) f(t)dt, differentiate wrt x, we get f'(x) = 0 + f'(x). This has the solution f(x)=e^x and x=1 gives e.

To take derivative, we used leibniz integral rule:

d/dx ( integral( a(x) to b(x) ) f(t) dt = f(b(x))*b'(x) - f(a(x))*a'(x) = f(x)*1 - f(x)*0 = f(x)

Otherwise, set up a (lag)differential equation for f(x), the expected number of draws needed for the sum to exceed x. For x=0,f(x)=1. For x>0, suppose a draw gave number t, then f(x)= 1 + integral (t=0 to x)f(x-t)dt, which is same as f(x) = 1 + integral(t=0 to x) f(t)dt, differentiate wrt x, we get f'(x) = 0 + f'(x). This has the solution f(x)=e^x and x=1 gives e.

To take derivative, we used leibniz integral rule:

d/dx ( integral( a(x) to b(x) ) f(t) dt = f(b(x))*b'(x) - f(a(x))*a'(x) = f(x)*1 - f(x)*0 = f(x)

Hard | Probability |

You are initially located at origin in the x-axis. You start a random walk with equal probability of moving left or right one step at a time. What is the probability that you will reach point a before reaching point -b? What is the expected number of steps to reach either a or -b? (a,b are natural numbers)

Answer

b/(a+b) and ab

Solution

If you try markov chain approach, this question will never end! For this problem, we will use Martingales.

Let N = number of steps to reach any of a or -b. Let SN = X1+...+XN. Suppose Pa = probability of touching a before -b, and Pb = the opposite. We use the fact that both SN and (SN^2-N) are martingales, thus:

0 = E(SN) = a*Pa + (-b)*Pb

This gives Pa = b/(a+b)

Similarly, 0 = E(SN^2-N) = Pa*a^2 + Pb*b^2 - E(N)

This gives E(N) = ab

Let N = number of steps to reach any of a or -b. Let SN = X1+...+XN. Suppose Pa = probability of touching a before -b, and Pb = the opposite. We use the fact that both SN and (SN^2-N) are martingales, thus:

0 = E(SN) = a*Pa + (-b)*Pb

This gives Pa = b/(a+b)

Similarly, 0 = E(SN^2-N) = Pa*a^2 + Pb*b^2 - E(N)

This gives E(N) = ab

Source: Top Quant Interview

Enable Like and Comment Hard | Probability |

You are taking out candies one by one from a jar that has 10 red candies, 20 blue candies, and 30 green candies in it. What is the probability that there are at least 1 blue candy and 1 green candy left in the jar when you have taken out all the red candies? (Candies of same color are indistinguishable!)

Hint

Combinatorial problem. Consider a sequence of 60 candies, in which red candies finish before blue and green.

Answer

7/12

Solution

It can be done using combinatorics as well as conditional probability. I don't have time today, but please remind me to put proper solution by commenting down below or by sending me a message. Thank you.

Hard | Strategy |

An aircraft hovers above sea, trying to catch a submarine moving with a constant velocity under the sea. The submarine is completely invisible, but using a human radar only once, the aircraft knows the exact location of submarine under the sea. The direction of submarine is unknown, but constant. The aircraft can move at twice the speed of submarine. As soon as the aircraft is just vertically above the submarine, Aircrafet can magnetically pick it up. How does the aircraft catch the submarine? How much time ill it take?

PS: This scene is from X-men: First Class. Good x-men are in the aircraft called blackbird, human radar is Banshee, Magnet is Magneto. Bad x-men are in submarine, with Sebastian Shaw is about to cause a war, better catch him soon!

PS: This scene is from X-men: First Class. Good x-men are in the aircraft called blackbird, human radar is Banshee, Magnet is Magneto. Bad x-men are in submarine, with Sebastian Shaw is about to cause a war, better catch him soon!

Hint

The locus of submarine from its spotted location is a circle of radius (speed of sub)*(time passed). We only need to move around this locus for some time.

Solution

Let speed of plane be 2s, submarine: s, original distance d

Locus of submarine after time t is circle of radius s*t centered at original location of submarine. Thus the plane moves 2/3 distance towards submarine, and then spirals out by increasing radius with speed 's'. The submarine is caught after one round.

For General case, with speeds p > s, see Palak's Answer:

Let Op, Os be the initial positions of the plane and the submarine, resp. For time t0 = d/(s+p), the plane will move towards Os in the OpOs direction. After that, with Os as the origin, the plane will maintain a constant velocity of s along the radial direction. Thus, at an angle theta from OpOs, tangential displacement in time dt is sqrt(p^2-s^2)*dt = r*d(theta) = s*t*d(theta) => dt/t = s/sqrt(p^2-s^2)*d(theta). Integrating t from t0 to tf, and theta from 0 to 2*pi, we get tf = t0*exp[2*pi*s/sqrt(p^2-s^2)] = [d/(s+p)]*exp[2*pi*s/sqrt(p^2-s^2)].

Note that the plane can be vertically above the submarine any time between t0 and tf, depending on the direction theta of the velocity of the submarine wrt OpOs, thus making tf the worst case time.

Locus of submarine after time t is circle of radius s*t centered at original location of submarine. Thus the plane moves 2/3 distance towards submarine, and then spirals out by increasing radius with speed 's'. The submarine is caught after one round.

For General case, with speeds p > s, see Palak's Answer:

Let Op, Os be the initial positions of the plane and the submarine, resp. For time t0 = d/(s+p), the plane will move towards Os in the OpOs direction. After that, with Os as the origin, the plane will maintain a constant velocity of s along the radial direction. Thus, at an angle theta from OpOs, tangential displacement in time dt is sqrt(p^2-s^2)*dt = r*d(theta) = s*t*d(theta) => dt/t = s/sqrt(p^2-s^2)*d(theta). Integrating t from t0 to tf, and theta from 0 to 2*pi, we get tf = t0*exp[2*pi*s/sqrt(p^2-s^2)] = [d/(s+p)]*exp[2*pi*s/sqrt(p^2-s^2)].

Note that the plane can be vertically above the submarine any time between t0 and tf, depending on the direction theta of the velocity of the submarine wrt OpOs, thus making tf the worst case time.

Source: Inspired from Rustan Leino's puzzle

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