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Chapter 6: Q. 6.15 (page 276)

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

Expert verified

The required probability is equal topk-1(1-p)n-kso it does not depend on the permutation of the sequence.

Step by step solution

01

Content Introduction

We are given,

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

02

Content Explanation

Suppose (Xn)nis that sequence of independent trial. we have Xn~Binom(p).

We are given Xn=1and in the random vector (X1,X2,.....,Xn-1)there exist K-1and others equal to zero. Therefore,

P(X1=x1,......,Xn-1=xn-1)=∏i=1n-1P(Xi=xi)=pk-1(1-p)n-k

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

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Most popular questions from this chapter

The “random” parts of the algorithm in Self-Test Problem 6.8 can be written in terms of the generated values of a sequence of independent uniform (0, 1) random variables, known as random numbers. With [x] defined as the largest integer less than or equal to x, the first step can be written as follows:

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