If I am given two repeatable events –
- event A – that I am told has a probability of 0.5% to produce result X
- event B – that I am told has a probability of 6% to produce result X
and given that the events have a cost associated with them, can I approximate how many times I would expect to have to perform event A to get result X with the same probability as performing event B once?
ie. how many times would I expect to have to do event A to get a probability of 6% of producing result X, to find out if the cost of doing event A that many times works out more or less expensive than just doing event B once?
Apologies if this has been asked before, but I don't know much about probability and what to search for, the other questions of this form all seem to want to know how many times to repeat event A to get a probability of 100% with some confidence level.
Best Answer
@Dave's answer is right, but I think you can approach it in a simpler way. Think instead of the probability of NOT getting result $X$, which is $0.995$ if you try once, $0.995^2$ if you try twice.... and $0.995^n$ if you try $n$ times.
In short, you only have to solve $0.995^n = 1-0.06 = 0.94$ With $n=13$ you get under $0.94$ for the first time (93.6% chance)
Finally, there is no way to guarantee that event $A$ will produce result $X$ for any number of trials, since $0.995^n$ is always greater than zero (it actually makes sense if you think about it)