Like, if something with a 10% chance happened 10 times, would it most likely happen once? same with 5% chance happening 1 in 20?
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A 10% chance is equal to a 1 in 10 chance. The rest is incorrect. I'll explain.
If something has a 1 in 10 chance of happening, it has a 1/10 chance, or 10%.
If you have a 10% chance of A happening and a 10% chance of B happening, assuming that the outcome of B is independent of the outcome of A, then the chance of both A and B happening is 10% times 10%, which is 1%. (Or, 1/10 times 1/10, which is 1/100).
The way to calculate the odds of something happening at least once is different. The chance of A not happening is 9/10 (the inverse of 1/10). The chance of B not happening is 9/10. The chance of A not happening and B not happening is 9/10 times 9/10, which is 81/100, or 81%.
You invert this to get a 19/100 (19%) chance of the inverse happening. What is the inverse of neither A nor B? It is A or B happening or both. Put another way, at least one "happening" out of the two tries.
What are the chances of at least one happening out of 10 tries if the chance is 10% each try? It is the same as above. The inverse of 9/10 multiplied by 9/10 ten times, which is 9/10 to the tenth power. When I do the math, the answer is that there is a 65.1% chance that something will happen at least once if you have 10 tries, with the chance of something happening in each try being 10%.
If you want to calculate this yourelf in Microsoft Excel, it is:
= 1 - ((1-1/10)^10)
If the chance of something happening is 1 in 20 (5%), then the chance of something happening at least once if you repeat this 20 times is 64.2%.
If something has a 1 in 10 chance of happening, it has a 1/10 chance, or 10%.
If you have a 10% chance of A happening and a 10% chance of B happening, assuming that the outcome of B is independent of the outcome of A, then the chance of both A and B happening is 10% times 10%, which is 1%. (Or, 1/10 times 1/10, which is 1/100).
The way to calculate the odds of something happening at least once is different. The chance of A not happening is 9/10 (the inverse of 1/10). The chance of B not happening is 9/10. The chance of A not happening and B not happening is 9/10 times 9/10, which is 81/100, or 81%.
You invert this to get a 19/100 (19%) chance of the inverse happening. What is the inverse of neither A nor B? It is A or B happening or both. Put another way, at least one "happening" out of the two tries.
What are the chances of at least one happening out of 10 tries if the chance is 10% each try? It is the same as above. The inverse of 9/10 multiplied by 9/10 ten times, which is 9/10 to the tenth power. When I do the math, the answer is that there is a 65.1% chance that something will happen at least once if you have 10 tries, with the chance of something happening in each try being 10%.
If you want to calculate this yourelf in Microsoft Excel, it is:
= 1 - ((1-1/10)^10)
If the chance of something happening is 1 in 20 (5%), then the chance of something happening at least once if you repeat this 20 times is 64.2%.