Calculate 8% Chance Event Occur After 3 Tries

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Discussion Overview

The discussion revolves around calculating the probability of an event with an 8% chance of occurring at least once after three independent trials. It explores different methods for determining this probability, including direct calculation and enumeration of outcomes.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant asks how to calculate the chance of an event occurring at least once after three tries, given an 8% chance per try.
  • Another participant questions the independence of trials and suggests writing a formula for the event not occurring at all in three tries.
  • A participant proposes that the probability of the event not occurring in three tries can be calculated as $$0.92^3$$, leading to the conclusion that the probability of it occurring at least once is $$1 - 0.92^3$$.
  • Another participant suggests a brute-force method to list all possible outcomes for three tries, calculating the probability for each outcome and summing those that meet the criterion.
  • A later reply expresses appreciation for the contributions made in the discussion.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the best method to calculate the probability, and multiple approaches are presented without resolution.

Contextual Notes

The discussion does not clarify the assumptions regarding the independence of trials or the completeness of the brute-force method proposed.

Drakkith
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An event has an 8% chance to occur per try. How would I calculate the chance of the event occurring at least once after 3 tries?
 
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Is the probability in one trial independent of the actual results in the others?

If so, can you write a formula for the event to not occur at all in three tries?
 
Drakkith said:
An event has an 8% chance to occur per try. How would I calculate the chance of the event occurring at least once after 3 tries?

The chance of it not occurring in 3 tries is $$0.92^3$$
So, you want $$1- 0.92^3$$
 
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To gain some insight into this problem, try the following brute-force method. List all the possible outcomes for the set of three tries:

1=no, 2=no, 3=no : probability = 0.92 * 0.92 * 0.92 = …
1=yes, 2=no, 3=no : probability = 0.08 * 0.92 * 0.92 = …
etc.

Enumerate all the outcomes (how many of them are there?), calculate the probability for each one, and add up the ones that meet your criterion.
 
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Awesome. Thanks guys.
 

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