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Consider a communication source that transmits packets containing digitized speech. After each transmission, the receiver sends a message indicating whether the transmission was successful or unsuccessful. If a transmission is unsuccessful, the packet is re-sent. Suppose a voice packet can be transmitted a maximum of \({\rm{10}}\) times. Assuming that the results of successive transmissions are independent of one another and that the probability of any particular transmission being successful is \({\rm{p}}\), determine the probability mass function of the rv \({\rm{X = }}\)the number of times a packet is transmitted. Then obtain an expression for the expected number of times a packet is transmitted.

Short Answer

Expert verified

The probability mass function is\({\rm{p(k) = P(X = k) = }}\left\{ {\begin{array}{*{20}{c}}{{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}{\rm{p}}}&{{\rm{k = 1,2,3,4,5,6,7,8,9}}}\\{{\rm{1 - }}\sum\limits_{{\rm{k = 0}}}^{\rm{9}} {{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}} {\rm{p}}}&{{\rm{k = 10}}}\end{array}} \right.\).

The expected number of times a packet is transmitted is \({\rm{\mu = 10 - 9}}\sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {{\rm{(1 - p)}}^{{\rm{k - 1}}}}{\rm{p}}\).

Step by step solution

01

Concept Introduction

Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring.

The complement rule is a statistical theorem that establishes a link between the probability of an occurrence and the probability of its complement, such that if one of these probabilities is known, automatically the other is also known.

02

Probability Mass Function

It is given that\({\rm{P(}}\)successful\({\rm{) = p}}\).

A packet can be transmitted at most\({\rm{10}}\)times –\({\rm{k}} \le {\rm{10}}\).

The transmissions are independent.

Probability Mass Function –

\({\rm{X = }}\)the number of times a packet is transmitted, thus\({\rm{X}}\)is the number of trials until the first success\({\rm{Y}}\)(successful transmission) occurs up to\({\rm{10}}\)packets.

The number of trials until the first success follows a geometric distribution with probability of success\(p\).

\({\rm{Y}} \sim {\rm{Geometric(p)}}\)

Definition geometric probability –

\({\rm{P(Y = k) = }}{{\rm{q}}^{{\rm{k - 1}}}}{\rm{p = (1 - p}}{{\rm{)}}^{{\rm{k - 1}}}}{\rm{p}}\)

The definition of geometric probability is for all nonnegative integers, however for \({\rm{X}}\) it is given \({\rm{k}} \le {\rm{10}}\). Use the definition for \({\rm{k = 1, 2, 3, 4, 5, 6, 7, 8, 9}}\).

03

The Complement Rule

The complement rule is represented as –

\({\rm{P(not A) = 1 - P(A)}}\)

Addition rule for disjoint or mutually exclusive events –

\({\rm{P(A or B) = P(A) + P(B)}}\)

Use the complement rule and the addition rule –

\(\begin{array}{c}{\rm{P(X = 10) = P(Y}} \ge {\rm{10) = 1 - P(Y}} \le {\rm{9) = 1 - P(Y = 1) - P(Y = 2) - \ldots - P(Y = 9)}}\\{\rm{ = 1 - }}\sum\limits_{{\rm{k = 0}}}^{\rm{9}} {\rm{P}} {\rm{(Y = k) = 1 - }}\sum\limits_{{\rm{k = 0}}}^{\rm{9}} {{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}} {\rm{p}}\end{array}\)

Combining these results to obtain the probability distribution –

\(\begin{array}{l}{\rm{p(k) = P(X = k)}}\\{\rm{ = }}\left\{ {\begin{array}{*{20}{c}}{{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}{\rm{p}}}&{{\rm{k = 1,2,3,4,5,6,7,8,9}}}\\{{\rm{1 - }}\sum\limits_{{\rm{k = 0}}}^{\rm{9}} {{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}} {\rm{p}}}&{{\rm{k = 10}}}\end{array}} \right.\end{array}\)

04

The Expected Value

The expected value (or mean) is the sum of the product of each possibility \({\rm{x}}\) with its probability \({\rm{P(x)}}\)–

\(\begin{aligned}\mu &= \sum\limits_{{\rm{10}}} {\rm{x}} {\rm{P(x)}}\\&= \sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {\rm{P(X = k)}}\\&= \sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {{\rm{(1 - p)}}^{{\rm{k - 1}}}}{\rm{p + 10}}\left( {{\rm{1 - }}\sum\limits_{{\rm{k = 0}}}^{\rm{9}} {{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}} {\rm{p}}} \right)\\&= \sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {{\rm{(1 - p)}}^{{\rm{k - 1}}}}{\rm{p + 10 - 10}}\sum\limits_{{\rm{k = 0}}}^{\rm{9}} {{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}} {\rm{p Distributive Property}}\\&= 10 + (1 - 10)\sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {{\rm{(1 - p)}}^{{\rm{k - 1}}}}{\rm{p Combine like terms}}\\&= 10 - 9\sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {{\rm{(1 - p)}}^{{\rm{k - 1}}}}{\rm{p}}\end{aligned}\)

Therefore, the expression is obtained as \({\rm{10 - 9}}\sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {{\rm{(1 - p)}}^{{\rm{k - 1}}}}{\rm{p}}\).

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Most popular questions from this chapter

Airlines sometimes overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets. Define the random variable Y as the number of ticketed passengers who actually show up for the flight. The probability mass function of Y appears in the accompanying table.

y

45

46

47

48

49

50

51

52

53

54

55

p(y)

.05

.10

.12

.14

.25

.17

.06

.05

.03

.02

.01

a. What is the probability that the flight will accommodateall ticketed passengers who show up?

b. What is the probability that not all ticketed passengerswho show up can be accommodated?

c. If you are the first person on the standby list (whichmeans you will be the first one to get on the plane ifthere are any seats available after all ticketed passengers have been accommodated), what is the probability that you will be able to take the flight? What is this probability if you are the third person on the standby list?

Let X have a Poisson distribution with parameter \({\rm{\mu }}\). Show that E(X) =\({\rm{\mu }}\) directly from the definition of expected value. (Hint: The first term in the sum equals \({\rm{0}}\), and then x can be cancelled. Now factor out \({\rm{\mu }}\) and show that what is left sums to \({\rm{1}}\).)

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday’s mail. In actuality, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one\(P\left( {Wed.} \right) = 0.3\), \(P\left( {Thurs.} \right) = 0.4\), \(P\left( {Fri.} \right) = 0.2\), and\(P\left( {Sat.} \right) = 0.1\). Let Y = the number of days beyond Wednesday that it takes for both magazines to arrive (so possible Y values are 0, 1, 2, or 3). Compute the pmf of Y. (Hint: There are 16 possible outcomes; \(Y\left( {W,W} \right) = {\bf{0}}\),\(Y\left( {F,Th} \right) = 2\), and so on.)

A personnel director interviewing \({\rm{11}}\) senior engineers for four job openings has scheduled six interviews for the first day and five for the second day of interviewing. Assume that the candidates are interviewed in random order. a. What is the probability that x of the top four candidates are interviewed on the first day? b. How many of the top four candidates can be expected to be interviewed on the first day?

A company that produces fine crystal knows from experience that 10% of its goblets have cosmetic flaws and must be classified as "seconds."

a. Among six randomly selected goblets, how likely is it that only one is a second?

b. Among six randomly selected goblets, what is the probability that at least two are seconds?

c. If goblets are examined one by one, what is the probability that at most five must be selected to find four that are not seconds?

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