# Probabilities with ${ n \choose k}$

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.

• Doesn't make sense. k is a parameter on the left and the index of summation on the right. Jan 20, 2020 at 2:48
• Note that s+f=1. Jan 20, 2020 at 4:30
• 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. Jan 20, 2020 at 4:39

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?