The probability of getting $k$ successes in $n$ trials

by SomeonecoolLovesMaths, Mar 20, 2024, 11:22 AM

The probability of getting $k$ successes in $n$ trials is $P_{r}={n\choose k}(p)^k(1-p)^{n-k}$ This is the binomial probability distribution. It has a finite number, $n$ , of trials. These repeated trials are independent of each other, so we multiply their probabilities. The outcome of each trial is deemed to be either a success, with probability $p$ , or a failure, with probability $q = 1 - p$ . The probability $p$ of a successful outcome is the same for each trial; and therefore the probability $1 - p$ of a failure is as well. There are
$\binom{n}{k}$
ways of exactly $k$ of the $n$ trials being successes and the probability of $k$ successes is $p^k$ . Then there are $n - k$ failures with probability $(1 - p)^{n - k}$ .
Thus the probability of getting exactly $k$ successes is
$\binom{n}{k} p^k (1-p)^{n-k}$ .

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The probability of getting $k$ successes in $n$ trials

by SomeonecoolLovesMaths, Mar 20, 2024, 11:22 AM

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