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Here is another way to obtain a set of recursive equations for determining Pn, the probability that there is a string of kconsecutive heads in a sequence of nflips of a fair coin that comes up heads with probability p:

(a) Argue that for k<n, there will be a string of kconsecutive heads if either

1. there is a string of kconsecutive heads within the first n-1flips, or

2. there is no string of kconsecutive heads within the first n-k-1flips, flip n-kis a tail, and flips n-k+1,…,nare all heads.

(b) Using the preceding, relate PntoPn-1. Starting with Pk=pk, the recursion can be used to obtain Pk+1, thenPk+2, and so on, up to Pn.

Short Answer

Expert verified

(a) 1 The probability for this case is Pn-1.

2 we have not obtained kconsecutive heads.

(b) Pn=Pn-1+(1-p)·1-Pn-k-1·pk

Step by step solution

01

Given information Part (a)  

We found that there is another way to obtain a set of recursive equations for determining Pn, the probability that there is a string of kconsecutive heads in a sequence of nflips of a fair coin that comes up heads with probability p.

02

 Explanation Part (a)  

Let's condition on what happens with the lastk flips. We have two options

1

There has been obtained kconsecutive Heads within first n-1flips. That means that we are not interested what happens in the last trial (the probability l), so the probability for this case is Pn-1.

2

Suppose that we have obtained kconsecutive Heads in the last kflips. That means that in n-k-1st flip we had to have tail and that in the first n-k-1flips we have not obtained kconsecutive heads.

03

Given information Part (b)  

We found that there is another way to obtain a set of recursive equations for determining Pn, the probability that there is a string of kconsecutive heads in a sequence of nflips of a fair coin that comes up heads with probability p.

04

Explanation Part (b)  

Considering the discussion in (a), the required recursion is

Pn=Pn-1+(1-p)·1-Pn-k-1·pk

That recursive relation can be solved using the method given in the exercise, but it is not part of this task.

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Most popular questions from this chapter

Aand Bwill take the same 10-question examination. Each question will be answered correctly by Awith probability.7, independently of her results on other questions. Each question will be answered correctly by B with probability .4, independently both of her results on the other questions and on the performance of A.

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