Hard probability question (cambridge exam question)

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Discussion Overview

The discussion revolves around a probability question from a Cambridge exam involving a character named Dipkomsky who uses multiple guns in a survival scenario. Participants are exploring the problem's requirements, including the preservation of a certain distribution and the calculation of expected value and variance related to the number of loaded guns.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion about how to start solving the problem and seeks assistance.
  • Another participant suggests sketching a probability tree for a single gun to aid understanding.
  • A participant discusses the case for m=1, analyzing the probabilities of a gun being loaded or not after the first round.
  • There is uncertainty about the meaning of "distribution" in the context of the problem, with questions about how it relates to the variable N.
  • Another participant indicates that deriving an expression for the probability after a move (p1) in terms of the initial probability (p) is necessary to show that p=3/7 is a solution.

Areas of Agreement / Disagreement

Participants generally agree that the problem is complex and requires careful analysis of probabilities, but there is no consensus on how to approach the solution or the implications of the distribution being preserved.

Contextual Notes

Participants note that understanding simpler probability tree questions may be beneficial for tackling this problem, indicating potential limitations in their current grasp of the concepts involved.

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original link: http://www.maths.cam.ac.uk/teaching/pastpapers/2001/Part_IA/PaperIA_2.pdf"

Question 11F

Dipkomsky, a desperado in the wild West, is surrounded by an enemy gang and
fighting tooth and nail for his survival. He has m guns, m > 1, pointing in different
directions and tries to use them in succession to give an impression that there are several
defenders. When he turns to a subsequent gun and discovers that the gun is loaded
he fires it with probability 1/2 and moves to the next one. Otherwise, i.e. when the
gun is unloaded, he loads it with probability 3/4 or simply moves to the next gun with
complementary probability 1/4. If he decides to load the gun he then fires it or not with
probability 1/2 and after that moves to the next gun anyway.
Initially, each gun had been loaded independently with probability p. Show that if
after each move this distribution is preserved, then p = 3/7. Calculate the expected value
EN and variance Var N of the number N of loaded guns under this distribution.

Hint: it may be helpful to represent N as a sum Xj (1 to m) of random variables
taking values 0 and 1.

This question is extremely confusing and I don't know even how to start, could anyone help?
 
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sweetpotatoes said:
This question is extremely confusing and I don't know even how to start, could anyone help?

To get started on the first part and to help your understanding, try sketching the probability tree for a single gun.

HTH
 
bpet said:
To get started on the first part and to help your understanding, try sketching the probability tree for a single gun.

HTH
bpet said:
To get started on the first part and to help your understanding, try sketching the probability tree for a single gun.

HTH

Thanks for the hints.
For m=1, then initially we have prob of p having it loaded, before the first round we have two case
loaded: prob of p
not loaded: 1-p
after 1st round
loaded:
not loaded: 1-p*1/4

Any idea what should I do next?

Also, I don't really understand what he means for "Show that if
after each move this distribution is preserved, then p = 3/7."
What exactly is the "distribution" referring to? is it N?
How could we use the property that the distribution is preserved

Thank you!
 
sweetpotatoes said:
... I don't really understand what he means for "Show that if after each move this distribution is preserved, then p = 3/7."

Say the probability after the move is p1, then p1=p. In the first part you'll have derived an expression for p1 in terms of p, and if your algebra is correct then p=3/7 is the only solution to the equation p1=p. To get the probabilities right though, you might need to practice on some simpler probability tree questions first.
 

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