# Derivation of Probability equation

#### masood

Hi All,
I need to understand derivation of the following equation, it is for particular network slot (tagged slot) which is in state $i$ (has currently $i$ users ) . The probability of having $\bar{D}=\bar{d}$ departures from it given that $D=d$ departures occur in all available slots including tagged slot is given as
$p_i( \bar{D}=\bar{d} | D=d ) = {i \choose \bar{d}} {N_t -i \choose d-\bar{d}} \pi_1^d (1-\pi_1)^{N_t-d}$ (given equation for derivation with $0 \leq \bar{d} \leq$ i ). $Nt$ is total number of users that can be using this slot.

It is also told that the number of departures from all slots including tagged slot is $d$. The departure of each users are independent events.

What i did for getting above result is shown below but i am not getting close to it.
since
$Pi( \bar{D}=\bar{d} | {D}={d} ) = \frac{ P (\bar{D}=\bar{d}, {D}={d}) } { P({D}={d}) }$
since departures are independent so I write it as

$Pi( \bar{D}=\bar{d} | {D}={d} ) = \frac{ P (\bar{D}=\bar{d}) P( {D}={d}) } { P({D}={d}) }$

it becomes

$Pi( \bar{D}=\bar{d} | {D}={d} ) = P (\bar{D}=\bar{d})$.

so i get following result by putting values in Bernoulli formula

$Pi( \bar{D}=\bar{d} | {D}={d} ) = \sum_{\bar{d}=0}^i {i \choose \bar{d}} p^{ \bar{d} } (i-p)^{ i-\bar{d} }$.

I thought that Bernoulli formula may be used here

$Pr[k\mbox{ successes in }n\mbox{ trials }] = \binom{n}{k}s^kf^{n-k}$.

or as

$Pr[k\mbox{ successes in }n\mbox{ trials }] =\sum_{k=0}^n \binom{n}{k}s^kf^{n-k}$ .

but how can convert $\sum$ into second ${ n \choose k }$ to derive given equation.

Am i right in making this decision ?. i am unable to derive it. Can any one tell me what i have done wrong?