Medium | Probability |

A very sharp, consistently skillful blind archer aimed for the center of a circular board and shot 2 arrows. He is expected to hit the aim, but doesn't hit it for sure. The archer is told that his first shot is better than second. He tried one more shot. What is the probability that this 3rd shot is the best shot among 3?

(ie, Probability that 3rd arrow lands closer to center than his first two shots?)

(ie, Probability that 3rd arrow lands closer to center than his first two shots?)

Hint

Enumeration; Symmetry between random variables.

Answer

1/3

Solution

Suppose x1, x2 and x3 are the distances of the arrows from center. As the archer is consistent, we can use symmetry, i.e., there are six equally likely cases: x1<x2<x3, (or 5 others, which are its permutations). Since archer is told that x1<x2, we are left with following equally likely cases:

x1<x2<x3 , x2<x1<x3 , x1<x3<x2

Among these three, one is favorable. Hence The probability of last shot being best is 1/3.

Notice that if there were (N-1) arrows, with first being better than rest, and then he shoots Nth arrow. The probability that Nth shot is best is 1/N.

x1<x2<x3 , x2<x1<x3 , x1<x3<x2

Among these three, one is favorable. Hence The probability of last shot being best is 1/3.

Notice that if there were (N-1) arrows, with first being better than rest, and then he shoots Nth arrow. The probability that Nth shot is best is 1/N.

Deadly | Probability |

Warning: I could not solve it.

In a room stand n armed and angry people. At each chime of a clock, everyone simultaneously spins around and shoots a random other person. The persons shot fall dead and the survivors spin and shoot again at the next chime. Eventually, either everyone is dead or there is a single survivor.

As n grows, what is the limiting probabality that there will be a survivor. :)

In a room stand n armed and angry people. At each chime of a clock, everyone simultaneously spins around and shoots a random other person. The persons shot fall dead and the survivors spin and shoot again at the next chime. Eventually, either everyone is dead or there is a single survivor.

As n grows, what is the limiting probabality that there will be a survivor. :)

Deadly | Probability |

The people in a country are partitioned into clans. In order to estimate the average size of a clan, a survey is conducted where 1000 randomly selected people are asked to state the size of the clan to which they belong. How does one compute an estimate average clan size from the data collected?

Solution

This is more of a puzzle-to-ponder rather than a puzzle to learn. In my opinion, best estimator for average is sum( n )/ sum( #n/n), where #n is the number of people with clan size as 'n', and this sum is over all the values of 'n' we receive.

Hard | Probability |

You are given an urn with 100 balls (50 black and 50 white). You pick balls from urn one by one without replacements until all the balls are out. A black followed by a white or a white followed by a black is "a colour change". Calculate the expected number of colour changes if the balls are being picked randomly from the urn.

Hint

Linearity of expectation

Solution

There are 99 positions. Let X_i be a random variable taking value 1 if i_th position has a colour change and zero otherwise.

We have to find expected value of E[X_1 + X_2 + ... + X_99]

Since all X_i are equivalent, the answer is 99*E[X_i]

E[X_i] = ((50/100)*(50/99)+(50/100)*(50/99)) = 50/99

So, Answer is 50.

We have to find expected value of E[X_1 + X_2 + ... + X_99]

Since all X_i are equivalent, the answer is 99*E[X_i]

E[X_i] = ((50/100)*(50/99)+(50/100)*(50/99)) = 50/99

So, Answer is 50.

Source: Placement test

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