Question

Consider a sequence of independent trials, with each trial being a success with probability p. Given that the kth success occurs on trial n, show that all possible outcomes of the first n − 1 trials that consist of k − 1 successes and n − k failures are equally likely.

Short answer

The required probability is equal to$p^{k-1}{(1-p)}^{n-k}$so it does not depend on the permutation of the sequence.

Step-by-step solution

Content Introduction

We are given,

Out of given independent trials, each trial has a probability of being success. Also,$X_n=1.$

Content Explanation

Suppose ${\left(X_n\right)}_n$is that sequence of independent trial. we have $X_n~Binom\left(p\right)$.

We are given $X_n=1$and in the random vector $(X_1,X_2,.....,X_{n-1})$there exist $K-1$and others equal to zero. Therefore,

$$\begin{gathered} P(X_1=x_1,......,X_{n-1}=x_{n-1})=\prod _{i=1}^{n-1}P(X_i=x_i) \\ =p^{k-1}{(1-p)}^{n-k} \end{gathered}$$

Since, the obtained sequence number does not depend on the permutation of the sequence, we have proved.