Independent trials that result in a success with probability $p$are successively performed until a total of $r$successes is obtained. Show that the probability that exactly$n$trials are required is

$\left(\begin{array}{l}n-1\\ r-1\end{array}\right)p^r(1-p{)}^{n-r}$

Use this result to solve the problem of the points

A set of trials conducted independently

In a set of experiments, the events S success and F failure occur.

$$\begin{gathered} P\left(S\right)=p, \\ \Rightarrow P\left(F\right)=1-p \end{gathered}$$

$A_{n,r}$- if n experiments are required to get r successes

Prove:

$P\left(A_{n,r}\right)=\left(\begin{array}{c}n-1\\ r-1\end{array}\right)p^r(1-p{)}^{n-r}$

Each order of successes and failures is a distinct occurrence from those defined by previous orders.

Because of their independence, the chances of a certain order of $r$succeeding in a $n$experiment are:

$P$(specific order of $r$successes and $n-r$failures) $=p^r(1-p{)}^{n-r}$

$A_{n,r}$is the result of combining a number of such events.

For a particular order, where the first $n$are the $r$successes

As a result, the nth trial must be a success for An, and the first $n-1$experiments must contain precisely $r-1$experiments. and each of the first n experiment outcomes is arranged in $A_{n,r}$.

There are $\left(\begin{array}{l}n-1\\ r-1\end{array}\right)$orders with $r-1$successes in each of the $n-1$experiments.

The union of $\left(\begin{array}{l}n-1\\ r-1\end{array}\right)$mutually excluded occurrences with probabilities of $n$is called $p^r(1-p{)}^{n-r}$.

$P\left(A_{n,r}\right)=\left(\begin{array}{c}n-1\\ r-1\end{array}\right)p^r(1-p{)}^{n-r}$