Calculate 8% Chance Event Occur After 3 Tries

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To calculate the probability of an event with an 8% chance occurring at least once after three tries, the formula used is 1 - (0.92^3), where 0.92 represents the probability of the event not occurring in a single trial. The independence of each trial means that the outcome of one does not affect the others. A brute-force method can also be employed by listing all possible outcomes for the three tries and calculating the probabilities for each. Ultimately, the total probability of the event occurring at least once can be derived from these calculations. This approach provides a clear understanding of the event's likelihood over multiple attempts.
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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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