Question

, or losing a data packet and packet losses are independent events A lost packet must be resert Rond your answers to fordernal places (·g. 98 7654). (a) What is the probability that an e-mail message with 100 packets will need any 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 are sent, each with 100 packets, what is the probability that at least one message wil need some packets to be resent

Answer 1

Let p=0.003 be the probability that any given data packet is lost.

a) Let X be the number of data packets that are lost in an e-mail message with 100 packets.

X has a Binomial distribution with parameters, number of trials n=100 and success probability (the probability that a data packet is lost) p=0.003

The probability that X=x data packets are lost in an e-mail message with 100 packets is given by the following Binomial pmf

$\begin{align*} P(X=x)&=\binom{n}{x}p^x(1-p)^{n-x}\\ &=\binom{100}{x}0.003^x(1-0.003)^{100-x}\\ &=\frac{100!}{x!(100-x)!}0.003^x(1-0.003)^{100-x}\\ \end{align*}$

The probability that an e-mail with 100 packets will need any resent, is same as the probability that an e-mail with 100 packets has at least 1 data packet lost (and hence needs at least one resent)

$\begin{align*} P(X\ge 1)&=1-P(X=0)\\ &=1-\binom{100}{0}0.003^0(1-0.003)^{100-0}\\ &=1-\frac{100!}{0!(100-0)!}0.003^0(1-0.003)^{100}\\ &=1-(1-0.003)^{100}\\ &=0.2595 \end{align*}$

ans: The probability that an e-mail with 100 packets will need any resent is 0.2595

b) Let Y be the number of data packets that are lost in an e-mail message with 3 packets.

Y has a Binomial distribution with parameters, number of trials n=3 and success probability (the probability that a data packet is lost) p=0.003

The probability that Y=y data packets are lost in an e-mail message with 3 packets is given by the following Binomial pmf

$\begin{align*} P(Y=y)&=\binom{n}{y}p^y(1-p)^{n-y}\\ &=\binom{3}{y}0.003^y(1-0.003)^{3-y}\\ &=\frac{3!}{y!(3-y)!}0.003^y(1-0.003)^{3-y}\\ \end{align*}$

The probability that an e-mail with 3 packets will need exactly 1 to be resent, is same as the probability that an e-mail with 3 packets has 1 data packet lost (and hence needs one resend)

$\begin{align*} P(Y=1)&=\binom{3}{1}0.003^1(1-0.003)^{3-1}\\ &=\frac{3!}{1!(3-1)!}0.003^1(1-0.003)^{3-1}\\ &=3\times 0.003^1(1-0.003)^2\\ &=0.0089 \end{align*}$

ans: The probability that an e-mail with 3 packets will need exactly 1 to be resent is 0.0089

c)We have calculated in part a) that  the probability that an e-mail with 100 packets will need some packages to be resent as p=0.2595

Let Z be the number of emails out of 10 emails with 100 packets that will need some packages to be resent.

Z has a Binomial distribution with parameters, number of trials n=10 and success probability (the probability that an  email will need some packages to be resent) p=0.2595

The probability that Z=z emails will need some packages to be resent is given by

$\begin{align*} P(Z=z)&=\binom{10}{z}0.2595^z(1-0.2595)^{10-z}\\ &=\frac{10!}{z!(10-z)!}0.2595^z(1-0.2595)^{10-z}\\ \end{align*}$

the probability that at least one message will need some packets to resent is

$\begin{align*} P(Z\ge 1)&=1-P(Z=0)\\ &=1-\binom{10}{0}0.2595^0(1-0.2595)^{10-0}\\ &=1-\frac{10!}{0!(10-0)!}0.2595^0(1-0.2595)^{10}\\ &=1-(1-0.2595)^{10}\\ &=0.9504 \end{align*}$

ans: the probability that at least one message will need some packets to resent is 0.9504