in 20 , in 25 , in 30 , etc. to a maximum of 75 trials ? ... or conversely , of NOT occurring in the same number or trials ? At what point will the event ALMOST certainly occur , or conversely, at what point will the event have only the REMOTEST possibility of NOT occurring ? Please suggest a formula for calculating other similar probabilities , e.g. , 1 in 8 , or 1 in 12 trials ... Sorry to say , that except for Standard Deviation , I've forgotten my math lessons of decades ago ...

Your response is much appreciated !

Quote:SuperpalPlease advise concerning the following question - If an event has a 1 in 10 probability of occurring , what is the probability of it occurring once in 15 trials ,

in 20 , in 25 , in 30 , etc. to a maximum of 75 trials ? ... or conversely , of NOT occurring in the same number or trials ? At what point will the event ALMOST certainly occur , or conversely, at what point will the event have only the REMOTEST possibility of NOT occurring ? Please suggest a formula for calculating other similar probabilities , e.g. , 1 in 8 , or 1 in 12 trials ... Sorry to say , that except for Standard Deviation , I've forgotten my math lessons of decades ago ...

Your response is much appreciated !

You can use a binomial distribution calculator which can be found online for free:

http://vassarstats.net/binomialX.html

You simply enter n as your number of attempts, k is how many of the desired outcome you want, and p is the probability it will occur given one chance.

Thus, for your example of fifteen trials: n=15, k=1, p=.1

For 1:15 attempts we get: 0.343151886824

For 0 or 1:15 attempts it is: 0.549043018919

TWENTY TRIALS:

1/20: 0.270170343535

0 or 1:20 0.391746998125

SEVENTY FIVE:

1:75 = 0.003083237375

0 or 1:75 0.00345322586

You can use this calculator to figure it out for anything you like, as long as there are only two possibilities for a result.

If it's "at least once", the probability = 1 minus the probability of it not happening at all, which is (1-p)

^{n}, where p is the probability of it happening in any particular trial (in your case, 0.1), and n is the number of trials.

If it's "exactly once", the probability = n times p times (1-p)

^{n-1}. ("n" is in there because any of the n trials can be the successful one.)

Quote:At what point will the event ALMOST certainly occur , or conversely, at what point will the event have only the REMOTEST possibility of NOT occurring ?

If you have the probability of an event occurring, p—then the probability that the event doesn’t occurs is 1-p. Then the probability of the event not occurring over n trials is (1-p)

^{n}. So, set this to whatever “remote” probability you want and solve for n.

So if p=0.1, and you want to know how many trials it takes for the event to occur 99% of the time (or to not occur 1% of the time)…

(1-0.1)

^{n}= 0.01

n log(0.9) = log(0.01)

n = 43.7

So it would take 44 trials to have over 99% confidence that something with 1 in 10 probability will happen at least once.

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Probability of occurring exact once in N trials = N * (0.1) * (0.9)^(N-1) = . . . . . .

You will get highest probability at N=9(or 10), probability = 9 * (0.1) * (0.9)^(9-1) = 0.387420489 . . .