Medium | Probability |

A coin is tossed 10 times and the output written as a string. What is the expected number of HH? Note that in HHH, number of HH = 2. (eg: expected number of HH in 2 tosses is 0.25, 3 tosses is 0.5)

Hint

Recursion

Solution

Let the expected number of HH for n tossed is E(n). So, probability that an (n-1) toss experiments, ends in T is 1/2.

So, E(n) = 1/2*E(n-1) + 1/2*( 1/2*(E(n-1)+1) + 1/2*(E(n-1)))

(The first case when it ends in T. & The second case when it ends in H.

In the second case, if you get an H then, you get 1 more HH. )

So, E(n) = E(n-1) + 1/4, us E(2) = 1/4

So, E(n) = (n-1)/4

For n=10, E(10) = 2.25

So, E(n) = 1/2*E(n-1) + 1/2*( 1/2*(E(n-1)+1) + 1/2*(E(n-1)))

(The first case when it ends in T. & The second case when it ends in H.

In the second case, if you get an H then, you get 1 more HH. )

So, E(n) = E(n-1) + 1/4, us E(2) = 1/4

So, E(n) = (n-1)/4

For n=10, E(10) = 2.25

Source: Placement Test

Enable Like and Comment Medium | Probability |

The stick drops and breaks at a random point distributed uniformly across the length. What is the expected length of the smaller part?

Solution

The probability that the break point belongs to first half is 1/2. In this case, the expected length of the small part is L/4. Its L/4 in the other case too. Hence E = 1/2*L/4 + 1/2*L/4 = L/4

Source: Quant Interview

Enable Like and Comment Hard | Probability |

There are 26 black(B) and 26 red(R) cards in a standard deck. A run is number of blocks of consecutive cards of the same color. For example, a sequence RRRRBBBRBRB of only 11 cards has 6 runs; namely, RRRR, BBB, R, B, R, B. Find the expected number of runs in a shuffled deck of cards.

Hint

Linearity of expectation

Answer

27

Solution

We form the answer for general deck of 2n cards, with n red cards and n black cards. Consider any sequence X1,X2..X2n (of R & B). Now define Y1 = 1;

Yi = 1 if Xi and X(i-1) are of different colors

Yi = 0 otherwise.

We want to find E[Y1 + ..+ Y2n], using linearity of expectation, = E(Y1) +...+E(Yn). Now note that, E[Y1] = 1 and E[Yi] = E[Yj] for the rest (by symmetry)

Also E[Yi] = E(Y2) = P(X2 is diff from X1) = (number of ways first two are RB or BR) / (Total number of orientations) = 2 * [ (2n-2)choose(n-1) ] / [ 2n choose n ] = n/(2n-1)

Ans = 1+(2n-1)*E[Y_i] = n+1

This is 27 for deck of 52 cards.

Yi = 1 if Xi and X(i-1) are of different colors

Yi = 0 otherwise.

We want to find E[Y1 + ..+ Y2n], using linearity of expectation, = E(Y1) +...+E(Yn). Now note that, E[Y1] = 1 and E[Yi] = E[Yj] for the rest (by symmetry)

Also E[Yi] = E(Y2) = P(X2 is diff from X1) = (number of ways first two are RB or BR) / (Total number of orientations) = 2 * [ (2n-2)choose(n-1) ] / [ 2n choose n ] = n/(2n-1)

Ans = 1+(2n-1)*E[Y_i] = n+1

This is 27 for deck of 52 cards.

Source: Quantnet

Enable Like and Comment Hard | Probability |

An ant is standing on one corner of a cube & can only walk on the edges. The ant is drunk and from any corner, it moves randomly by choosing any edge! What is the expected number of edges the ant travels, to reach the opposite corner?

Hint

Try to find the equivalent vertices with respect to distance yet to travel. This should give 4 equivalent merged vertices, with 1st being start & 4th being destination.

Solution

Let the expected number of step required to go from (0,0,0) to (1,1,1) be E0.Also let expected number of step required to reach (1,1,1) from (0,0,1)(Or from (0,1,0) or from (1,0,0)) be E1. similarly expected number of step required to reach (1,1,1) from (0,1,1) (Or from (1,0,1) or from (1,1,0)) be E2.Then we can write:

E0=1/3(E1+E1+E1)+1

E1= 1/3*E0 + 2/3*E2 + 1

E2 = 2/3*E1 + 1

solving this we find E0 as 10.

E0=1/3(E1+E1+E1)+1

E1= 1/3*E0 + 2/3*E2 + 1

E2 = 2/3*E1 + 1

solving this we find E0 as 10.

Source: Quant Interview

Enable Like and Comment Hard | Probability |

What is the expected distance of any point on Earth and the north pole? Take Earth radius 1.

Clarification: Shortest distance cuts through the sphere, instead of lying on surface.

Further thinking: Is this question same as choosing two random points on unit sphere and asking their expected distance?

Clarification: Shortest distance cuts through the sphere, instead of lying on surface.

Further thinking: Is this question same as choosing two random points on unit sphere and asking their expected distance?

Hint

Imagine a ring of some thickness, whose distance from N is constant from all points on that ring. Get the average of all such rings.

Answer

4/3

Solution

2*sin(x/2)is the distance of north pole from a point on the ring at angle x from the z axis. So, integrate from 0 to pi, (2*pi*sin(x)*2*sin(x/2) dx) and divide by the total area which is 4*pi.

Answer:4/3

Another approach is to imagine a horizontal ring of dy thickness at distance y from N (north pole).

Area of ring = 2*pi*dy

Probability of choosing point on this ring = dy/2

Distance of N & a point on ring = sqrt(2y)

Exp length = integral (y=0 to 2) of sqrt(2y) dy/2 = 4/3

And yes, taking two random points on surface of sphere and asking their expected distance is same as this very question.

Answer:4/3

Another approach is to imagine a horizontal ring of dy thickness at distance y from N (north pole).

Area of ring = 2*pi*dy

Probability of choosing point on this ring = dy/2

Distance of N & a point on ring = sqrt(2y)

Exp length = integral (y=0 to 2) of sqrt(2y) dy/2 = 4/3

And yes, taking two random points on surface of sphere and asking their expected distance is same as this very question.

Source: Quant Interview

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