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I'm looking for chance of success when within a number of trials with each trial having success rate x

I learned that formula in highschool stats but I've since forgotten it. Oh what a fool am I!

My intuition says that: 50 x 2 = 100, 10 x 2 = 20

so 20%

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  • $\begingroup$ you might look into the problem this way: each trial has a failure rate of 98%. We assume trials are independent, so the $P($all fail$)=0.98^{10}$ and the probability that at least one of them succeeds is $1-P($ all fail$)$. If you just want 1 to succeed and the others to fail you can model this as a biased coin flip with $p=0.02$. It is ${ 10 \choose 1}p(1-p)^9$ $\endgroup$ – V. Aslanyan Sep 26 '18 at 17:00
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    $\begingroup$ I would also suggest editing the question since it is unclear which probability you are looking for $\endgroup$ – V. Aslanyan Sep 26 '18 at 17:02
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    $\begingroup$ Your intuition is good but you need to understand it has limited scope. For instance, if there were 100 trials, what result would you get? :-) $\endgroup$ – whuber Sep 26 '18 at 17:33
  • $\begingroup$ 200% aka 2 successes? $\endgroup$ – bmoriciv Sep 26 '18 at 17:38
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    $\begingroup$ @bmoriciv so you are specifically looking for 1 success only. Intuition behind my thinking is: we want 1 experiment to be success and the other to fail. In addition, we don't care which of our 10 trials succeeds, so we add all 10 possibilities together. thus we get $p(1-p)^9$ for the first trial to succeed, $p(1-p)^9$ for the second and so on. Thus, we have $10p(1-p)^9$ for total probability of one success and 9 failures $\endgroup$ – V. Aslanyan Sep 26 '18 at 17:46
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You need to review and use the binomial distribution which gives you the probability of a certain number of successes in a certain number of trials.

Check the wikipedia example:

https://en.wikipedia.org/wiki/Binomial_distribution#Example

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