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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