Probability of an event is $.116$. In $50$ tries, what are the chances at least one event occurs? I see that the probability that it wouldn't happen in one try is $.884$ and the probability that it wouldn't happen in two tries is $(.884)^2$.
$\endgroup$43 Answers
$\begingroup$If an event occurs $0$ times (out of $50$, in this case) then it does not occur at least once. So we can find the probability of it not occurring and then subtract that value from $1$.
So, what are the chances of it not occurring on $1$ trial? $1-.116 = .884$
What about not occurring on $2$ trials? $(1-.116)\cdot (1-.116)=.884 \cdot .884 = .781456$
Now what about not occurring on $50$ trials? $.884^{50} \approx .002102$
We must subtract this value from $1$ (recall that what we just calculated is the probability of it not occurring, so the probability of it occuring at least once is: $$1-.002102 \approx .9979 $$
$\endgroup$$\begingroup$A more general approach to this problem is through the binomial distribution.
If $p$ is the probability of success of an event, then the probability that we have exactly $k$ successes over $n$ independent trials is: $$P(X = k) = \dbinom{n}{k} \cdot p^k \cdot (1-p)^{n-k},$$ where $X$ is the random variable that expresses the number of successes of the event.
In our specific problem, we have that $p= 0.116, n = 50$ and we want $$P(X\ge 1) = P(X = 1) + P(X = 2)+ \cdots + P(X = 50) = 1-P(X = 0)$$
$\endgroup$$\begingroup$The chances of the event not happening after 1 trial is (1-0.116) = 0.884
The chances of the event not happening after 2 trials is $0.884^2$
The chances of the event not happening after 50 trials is
0.002101905........... ($0.884^{50}$)
Therefore, the chances of the event happening at least once is 1 minus the above number which is
0.99789880941.........
There is a 99.79% (to 4sf) chance the event will happen at least once in 50 trials.
$\endgroup$
