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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

NBC News reported on May 2,2013, that 1 in 20 children in the United States have a food allergy of some sort. Consider selecting a random sample of 25 children and let X be the number in the sample who have a food allergy. Then \(X~Bin (25,.05)\).

a. Determine both \(P(X \le 3)\)and \(P(X < 3)\).

b. Determine \(P(X \ge 4)\).

c. Determine \(P(1 \le X \le 3)\).

d. What are E(X) and \({\sigma _X}\)?

e. In a sample of 50 children, what is the probability that none has a food allergy?

A family decides to have children until it has three children of the same gender. Assuming P(B) = P(G) =\({\rm{.5}}\), what is the pmf of X = the number of children in the family?

A new battery’s voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that 90% of all batteries have acceptable voltages. Let Y denote the number of batteries that must be tested.

a. What is\(p\left( 2 \right)\), that is, \(P\left( {Y = 2} \right)\),?

b. What is\(p\left( 2 \right)\)? (Hint: There are two different outcomes that result in\(Y = 3\).)

c. To have \(Y = 5\), what must be true of the fifth battery selected? List the four outcomes for which Y = 5 and then determine\(p\left( 5 \right)\).

d. Use the pattern in your answers for parts (a)–(c) to obtain a general formula \(p\left( y \right)\).

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.)

Customers at a gas station pay with a credit card (A), debit card (B), or cash (C). Assume that successive customers make independent choices, with P(A)\({\rm{ = }}{\rm{.5}}\), P(B)\({\rm{ = }}{\rm{.2}}\), and P(C)\({\rm{ = }}{\rm{.3}}\). a. Among the next\({\rm{100}}\)customers, what are the mean and variance of the number who pay with a debit card? Explain your reasoning. b. Answer part (a) for the number among the\({\rm{100}}\)who don’t pay with cash.

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