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Probability & Statistics/Probability chance over multiple probability events

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Question
If an event has a 10% chance of an outcome, what is the percentage chance that outcome will have occurred if the event happens 10 times? More importantly, how is this calculated?

Probability tables donít seem to help because this leaves you with 10 in 100, which still only figures what would happen in one particular event, not the sum probability of ten separate events. I tried one method of taking .1 x .5 (too say the event happened again) and taking that number, .05, x .5 and so on up to 10 and then adding up all the amounts, but that left me with a figure that would never exceed 20%, which isnít right.

This should be an easy thing to figure out, I'm just not that math savvy!

Thanks.

Answer
If an event has probability p, then the probability that it happens at least once in k trials can be computed by:

1 - (1-p)^k

In your case, the answer is 1 - (1-1/10)^10 ≈ 65%

This, translated into words, sort of says "Not Never The Event" -- how?

Step 1: p is the event (call it E)
Step 2: 1-p is "not E"
Step 3: (1-p)^k is "never E in k trials"
Step 4: 1-(1-p)^k is "not never E in k trials" which is the same as "at least once E in k trials"


Semantics aside, there are only three concepts here:

Concept 1. Complementary probabilities (the events E and "not E")

If p is the probability of the event E, then 1-p is the probability of E not happening.

The same goes for (1-p)^k being the opposite or "complement" of 1-(1-p)^k.

Concept 2. "Not never" means "at least once"

This concept is pretty much self-explanatory.

Concept 3. Multiple trials are exponents, like p^k or (1-p)^k

If you have some trial and an event F (maybe not our event "E"), you want to figure out what are the odds that F occurs every single time. Like if you want to win the lottery twice, and p is the probability that you win the lottery once, then the probability is p^2.

This rests on the assumption that the trials are independent. If you win the lottery, you are not especially likely to have a winning ticket again -- the lottery machine is still random and independent of previous outcomes.

So if you wanted to have your event E occur all 10 times, you need p^k, where k is the number of trials and p is the probability of E. That would be (1/10)^10.

However, that is not what you want. You only want E to occur once (or more maybe). What this really means is that there is only one excluded outcome -- that all ten trials go, and they all fail to produce E. In that case, the event you want to look at is "E never happens in k trials" which is (1-p)^k, for you that is (1-1/10)^10. This is the only outcome (of all k trials) that is excluded, giving the answer:

1 - (1-p)^k = 1 - (1 - 1/10)^10

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Clyde Oliver

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I can answer all questions up to, and including, graduate level mathematics. I do not have expertise in statistics (I can answer questions about the mathematical foundations of statistics). I am very much proficient in probability. I am not inclined to answer questions that appear to be homework, nor questions that are not meaningful or advanced in any way.

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I am a PhD educated mathematician working in research at a major university.

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AMS

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BA mathematics & physics, PhD mathematics from a top 20 US school.

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Various honors related to grades, various fellowships & scholarships, awards for contributions to mathematics and education at my schools, etc.

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In the past, and as my career progresses, I have worked and continue to work as an educator and mentor to students of varying age levels, skill levels, and educational levels.

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