Question

Consider n independent sequential trials, each of which is successful with probability p. If there is a total of k successes, show that each of the n!/[k!(n − k)!] possible arrangements of the k successes and n − k failures is equally likely.

Short answer

The answer is$p=\frac{p^k·(1-p{)}^{n-k}}{\left(\begin{array}{l}n\\ k\end{array}\right)p^k(1-p{)}^{n-k}}=\frac{1}{\left(\begin{array}{l}n\\ k\end{array}\right)}$.

Step-by-step solution

Step 1:Given Information

We have that the probability the probability for $k$successes in $n$trials is simply $\left(\begin{array}{l}n\\ k\end{array}\right)p^k(1-p{)}^{n-k}$.On the other hand ,the probability for some certain arrangement is $p^k·(1-p{)}^{n-k}$.

Step 2:Explanation

The probability for some certain arrangement given that we had$k$successes in$n$trials is simply$p=\frac{p^k·(1-p{)}^{n-k}}{\left(\begin{array}{l}n\\ k\end{array}\right)p^k(1-p{)}^{n-k}}=\frac{1}{\left(\begin{array}{l}n\\ k\end{array}\right)}$.

Step 3:Final Answer

The answer is$p=\frac{p^k·(1-p{)}^{n-k}}{\left(\begin{array}{l}n\\ k\end{array}\right)p^k(1-p{)}^{n-k}}=\frac{1}{\left(\begin{array}{l}n\\ k\end{array}\right)}$