Suppose that in Problem 7.70, we continue to flip the coin until a head appears. Let Ndenote the number of flips needed. Find

(a)P{Ni},i1

(b)P{N=i};

(c)E[N]

Short Answer

Expert verified

a) The value of P{Ni},i1is P[Ni]=1i;i=0,1,2,.,n

b) The value of P{N=i}is P[N=i]=1(i)(i+1);i=0,1,2,,n

c) The value ofE[N]is.

Step by step solution

01

Given Information (Part a)

Flip the coin until a head appears.

Number of flips needed =N

P{NI},i1=?

02

Explanation (Part a) 

We have,

P[N=i]=01i10p(1p)i1dp

=01p(1p)i1dp

localid="1647430523751" =(1)!(i1)!(2+i1)!

localid="1647429441069" =(i1)!(i+1)!

=1i(i+1)

P[Ni]=x=i1x(x+1)

localid="1647429510678" =x=i1x1x+1

=1i

HenceP[Ni]=1i;i=0,1,2,.,n

03

Final Answer

Hence, the value ofP{Ni},i1isP[Ni]=1i;i=0,1,2,,n

04

Given Information (Part b)

Flip the coin until a head appears.

Number of flips needed=N

P{N=i}=?

05

Explanation (Part b) 

We have,

P[N=i]=P[Ni]P[N>i]

=P[Ni]P[Ni+1]

=1i1i+1

localid="1647429927042" =1i(i+1)

P[N=i]=1(i)(i+1);i=0,1,2,.,n

06

Final Answer (Part b) 

Hence, the value of E[N]is.

07

Given Information (Part c)

Flip the coin until a head appears.

Number of flips needed =N

The Value ofE[N]=?

08

Explanation (Part c) 

E(N)=i=0P[Ni]

role="math" localid="1647431080445" =i=01i

=

09

Final Answer (Part c) 

Therefore, the value ofE[N]=

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

LetX have moment generating function M(t), and defineΨ(t)=logM(t). Show thatΨ''(t)t=0=Var(X).

The number of accidents that a person has in a given year is a Poisson random variable with mean λ̣ However, suppose that the value of λchanges from person to person, being equal to 2for 60percent of the population and 3for the other 40percent. If a person is chosen at random, what is the probability that he will have

(a) 0accidents and,

(b) Exactly 3accidents in a certain year? What is the conditional probability that he will have3 accidents in a given year, given that he had no accidents the preceding year?

In an urn containing n balls, the ith ball has weight W(i),i = 1,...,n. The balls are removed without replacement, one at a time, according to the following rule: At each selection, the probability that a given ball in the urn is chosen is equal to its weight divided by the sum of the weights remaining in the urn. For instance, if at some time i1,...,ir is the set of balls remaining in the urn, then the next selection will be ij with probability W(ij)/k=1rW(ik), j = 1,...,r Compute the expected number of balls that are withdrawn before the ball number 1is removed.

A bottle initially contains m large pills and n small pills. Each day, a patient randomly chooses one of the pills. If a small pill is chosen, then that pill is eaten. If a large pill is chosen, then the pill is broken in two; one part is returned to the bottle (and is now considered a small pill) and the other part is then eaten.

(a) Let X denote the number of small pills in the bottle after the last large pill has been chosen and its smaller half returned. Find E[X].

Hint: Define n + m indicator variables, one for each of the small pills initially present and one for each of the small pills created when a large one is split in two. Now use the argument of Example 2m.

(b) Let Y denote the day on which the last large pills chosen. Find E[Y].

Hint: What is the relationship between X and Y?

If Y=aX+bwhere a and b are constants, express the moment generating function of Y in terms of the moment generating function of X.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free