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The probability of getting $k$ successes in $n$ trials is given by

$$P(k)={n \choose k} p^k q^{n-k}$$

where the probability of a particular outcome (a string made of $1$s and $0$s of length $n$, $1$- success, $0$- failure) with $k$ successes equals $p^k q^{n-k}$.

Why do we multiply $p^k q^{n-k}$ by ${n \choose k}$? I know there are ${n \choose k}$ such distinct outcomes, but from the point of view of probability -

Is it because we treat every particular outcome as a separate event (it has to be the case, because probability function assigns probabilities to events, every events is a collection of outcomes, and in this case each event consists of exactly one outcome), and because these events are disjoint, we can add up the probabilities?

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    $\begingroup$ $p^k q^{n-k}$ is the probability of one specific arrangement of the one's among the zeros. There are ${n\choose k}$ such different arrangements. We treat them as separate distinct events. If we had to count those cases when there were $k$ successes then we wouldn't want to handle the the different possibilities separately. (As you explained quite well.) $\endgroup$ – zoli Dec 13 '15 at 10:36
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You can get exactly $k$ successes (and $n-k$ failures) in $n$ trials ($k\le n$) in many ways. For example you can succeed in the first $k$ trials or in the last $k$ trials or in the first trial, fail in the second and then succeed in the subsequent $k-1$ trials or $\ldots$ etc etc. How many ways are there? Exactly $\dbinom{n}{k}$. Why did I highlight the "or's"? Because "or" in events translates to addition of the corresponding probabilities. So, the total probability is given if you add $\dbinom{n}{k}$ times the probability $p^k(1-p)^{n-k}$.

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