can i convert $\sum_{k=0}^n$ in to ${a \choose b}$ form in the Bernoulli Equation shown below:

$Pr[k\mbox{ successes in }n\mbox{ trials }] =\sum_{k=0}^n \binom{n}{k}s^kf^{n-k}$ , $s$ and $f$ are probabilities of success and failure respectively.

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    $\begingroup$ Doesn't make sense. k is a parameter on the left and the index of summation on the right. $\endgroup$ Jan 20, 2020 at 2:48
  • $\begingroup$ Note that s+f=1. $\endgroup$ Jan 20, 2020 at 4:30
  • $\begingroup$ Thanks for the reply, actually i have posted this question but no one replied math.stackexchange.com/questions/3514268/… Here i refere to equation that has this logic. please see this. i will be very thankful. $\endgroup$ Jan 20, 2020 at 4:39

1 Answer 1


Actually I need help deriving this equation

$p_i( \bar{D}=\bar{d} | D=d ) = {i \choose \bar{d}} {N_t -i \choose d-\bar{d}} \pi_1^d (1-\pi_1)^{N_t-d}$ (given equation for derivation with $ 0 \leq \bar{d} \leq$ i ). As number of departure from tagged slot cannot be greater than number of users using it)

The given equation is w.r.t a particular slot (tagged slot) which is in state $i$ (has currently $i$ users ) and the number of departure of users from this slot is termed as $\bar{d}$. The number of departures from all all slots including tagged slot is $d$. The departure of each users are independent events with $Nt$ as total number of users.

What i think is that 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?


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