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+ | Bits are sent over a communications channel in packets of 12. If | ||
+ | the probability of a bit being corrupted over this channel is 0.1 and | ||
+ | such errors are independent, what is the probability that no more | ||
+ | than 2 bits in a packet are corrupted? | ||
+ | If 6 packets are sent over the channel, what is the probability that | ||
+ | at least one packet will contain 3 or more corrupted bits? | ||
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Latest revision as of 16:28, 23 February 2017
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2013
Part 1
Consider $ n $ independent flips of a coin having probability $ p $ of landing on heads. Say that a changeover occurs whenever an outcome differs from the one preceding it. For instance, if $ n=5 $ and the sequence $ HHTHT $ is observed, then there are 3 changeovers. Find the expected number of changeovers for $ n $ flips. Hint: Express the number of changeovers as a sum of Bernoulli random variables.
Solution 1
The number of changeovers $ Y $ can be expressed as the sum of n-1 Bernoulli random variables:
$ Y=\sum_{i=1}^{n-1}X_i $.
Therefore,
$ E(Y)=E(\sum_{i=1}^{n-1}X_i)=\sum_{i=1}^{n-1}E(X_i) $.
For Bernoulli random variables,
$ E(X_i)=p(E_i=1)=p(1-p)+(1-p)p=2p(1-p) $.
Thus
$ E(Y)=2(n-1)p(1-p) $.
Comments on Solution 1:
Good solution with appropriate explanation.
Solution 2
For n flips, there are n-1 changeovers at most. Assume random variable $ k_i $ for changeover,
$ P(k_i=1)=p(1-p)+(1-p)p=2p(1-p) $
$ E(k)=\sum_{i=1}^{n-1}P(k_i=1)=2(n-1)p(1-p) $
Critique on Solution 2:
The solution is correct. However, it's better to explicitly express $ k_i $ as a Bernoulli random variable. This makes it easier for readers to understand.
Solution 3
First, we define a Bernoulli random variable
$ X = \left\{ \begin{array}{ll} 0, & the change over does not occur\\ 1, & the change over occurs \end{array} \right. $
Then we can compute
$P(X = 1) = P(1-P)+(1-P)P = P-{P}^{2}+P-{P}^2=2P-2{P}^{2} $
$P(X = 0) = P•P+(1-P)(1-P) = {P}^{2}+1-2P+{P}^2 $
Define Y as the number of changes occurred in n flips, there exists at most n-1 changes
$E(Y)=E(\sum_{i=1}^{n-1}X_i)=\sum_{i=1}^{n-1}E(X_i) $
$E(X_i)=p(X_i=1)=p(1-p)+(1-p)p=2p(1-p) $
Therefore, we have a final solution as
$ E(Y)=2(n-1)p(1-p) $.
Similar Question
Bits are sent over a communications channel in packets of 12. If the probability of a bit being corrupted over this channel is 0.1 and such errors are independent, what is the probability that no more than 2 bits in a packet are corrupted? If 6 packets are sent over the channel, what is the probability that at least one packet will contain 3 or more corrupted bits?