1
$\begingroup$

this is what is in my probability book: Let $X$ be the number of independent Bernoulli trials, each with success probability $p$, up to and including the $r$th success. $X$ is a discrete random variable taking values in $\{ r, r+1,... \}$. Then

$$P(X=k) = P(r-1 \text{ successes in first } k-1 \text{ throws}, 1 \text{ success} \text{ on } k\text{th throw}) = {k-1 \choose r-1}p^{r-1}q^{k-r}\times p = {k-1 \choose r-1} p^rq^{k-r}$$ for $k = r,r+1,...$

I'm not understand how this is derived. I thought the negative binomial is the generalised geometric distribution. The geometric distribution is where $X$ is the number of trials up to and including the $kth$ success, so I'm not really sure what the negative binomial distribution is, and what the author means by $k-1$ "throws". Any explanation please

$\endgroup$
0
$\begingroup$

I dont get the question very well. But I can tell you what

$${k-1 \choose r-1}p^{r-1}q^{k-r}\times p = {k-1 \choose r-1} p^rq^{k-r}$$

means:

You create a stream of Binary numbers, each having a success probability $p$ and failure probability $q=1-p$. The places of $1$s can be configured in $r-1$ different ways out of total $k-1$ binary numbers. For example if $k-1=2$ then

$00$, $01$, $10$ and $11$ are all possible configurations and if there are say $r-1=1$ success then $$ \binom{k-1}{r-1}=\binom{2}{1}=2$$

which indicates the bit streams $01$, $10$: one success out of two.

It seems that every stream that is selected has a probability of $p$ to happen and this explains the given formula.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.