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A congested computer network has a 0.002 probability of losing a data packet and packet losses are independent events. A lost packet must be resent. (a) What is the probability that an e-mail message with 100 packets will need any to be resent? (b) What is the probability that an e-mail message with 3 packets will need exactly one to be resent? (c) If 10 e-mail messages are sent, each with 100 packets, what is the probability that at least one message will need some packets to be resent?

Short Answer

Expert verified
(a) Approximately 0.181; (b) Approximately 0.006; (c) Approximately 0.840.

Step by step solution

01

Understanding the Problem

We're dealing with a problem involving probability of packet losses in a computer network. We have a probability of 0.002 for losing a single packet, and there are 100 packets in an email message. We need to compute different probabilities concerning packets being resent.
02

Probability of Not Losing Any Packet (100 packets)

First, we find the probability of not losing any packet in a message of 100 packets. The probability of not losing a single packet is 1 - 0.002 = 0.998. Since packet losses are independent, the probability of not losing any packets is \( (0.998)^{100} \).
03

Probability at Least One Packet Lost (100 packets)

Now, we find the probability that at least one packet is lost in a message of 100 packets, which is the complement of not losing any packet: \( 1 - (0.998)^{100} \).
04

Calculation for Part (a)

Calculate \( 1 - (0.998)^{100} \) to find the probability that at least one packet needs to be resent in a message of 100 packets.
05

Probability of Exactly One Packet Lost (3 packets)

For an email message with 3 packets, we want the probability of exactly one packet being lost. This follows a binomial distribution: \( P(X = 1) = \binom{3}{1} \cdot (0.002)^1 \cdot (0.998)^{2} \), where \( \binom{3}{1} \) is the number of ways to choose 1 lost packet out of 3.
06

Calculation for Part (b)

Calculate \( 3 \cdot (0.002) \cdot (0.998)^2 \) to find the probability that exactly one packet is lost in a message of 3 packets.
07

At Least One Message Needs Resending (10 messages)

We have 10 email messages each with 100 packets. To find the probability that at least one message needs a packet resent, first find the probability that none of the messages need resending, then subtract from 1. For one message, the probability of not resending is \( (0.998)^{100} \); for 10 messages, it's \( ((0.998)^{100})^{10} \).
08

Calculation for Part (c)

Finally, calculate \( 1 - ((0.998)^{100})^{10} \) to find the probability that at least one of 10 messages needs some packets to be resent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independent Events
In probability theory, independent events are those whose occurrence or non-occurrence does not affect the probability of each other. Understanding this concept is crucial when dealing with scenarios like data packet transmission, where the loss of one packet does not influence the loss of another.
To illustrate, suppose the probability of losing a single packet in a computer network is 0.002. Because packet losses occur independently, if there are 100 packets in a message, the probability of not losing any packet can be calculated by raising the probability of not losing a single packet to the power of the total number of packets.
This is expressed mathematically as \((1 - 0.002)^{100}\), which reflects the nature of independent events. The concept simplifies calculations because we can multiply probabilities directly when they are independent. It's an essential principle when we need to compute the likelihood of complex events occurring.
Binomial Distribution
The binomial distribution is a fundamental probability distribution that models the number of successes in a fixed number of trials. Each trial is independent and has the same probability of success. It's particularly applicable when you're dealing with packet losses over a network, where each packet transmission can be considered a trial.
For instance, if you want to calculate the probability of exactly one packet loss in a message containing 3 packets, you use the binomial distribution formula. The formula is \( P(X = k) = \binom{n}{k} \, p^k \, (1-p)^{n-k} \), where \( n \) is the total number of packets, \( k \) is the number of packet losses (successes in this context), \( p \) is the probability of losing a packet, and \( \binom{n}{k} \) is the binomial coefficient.
  • \( n = 3 \)
  • \( k = 1 \)
  • \( p = 0.002 \)
Plug these values into the formula to find the probability of exactly one packet being lost.
Complementary Probability
Complementary probability is a handy concept for calculating the likelihood of an event not happening. It's based on the understanding that the probability of an event occurring and not occurring must total to 1. This is especially useful when you're more interested in the opposite of the event you're analyzing.
For example, in our packet loss problem, you might be tasked with finding the probability that at least one packet is lost. You could calculate this directly, but it's often easier to first compute the probability of losing no packets at all, using the formula \((0.998)^{100}\), and then subtracting this result from 1.
This technique greatly simplifies the calculation, making it quicker and more intuitive to solve problems involving events that occur infrequently, such as packet loss in a large number of transmissions. By focusing on the complementary event, you can navigate complex probability problems with greater ease.

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