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Bits are sent through an information channel where each bit has $0.1$ probability of being transferred incorrectly, independent from other bits. (for example $0$ is sent and $1$ is received).

Bits will be sent in sequence consisting of $5$ bits. Bits from this sequence will be decoded according to the most frequent bit in sequence

Examples

Sent $\rightarrow$ Recieved

$00000 \rightarrow 0$

$01010 \rightarrow 0$

$01101 \rightarrow 1$

$11111 \rightarrow 1$

Questions:

(a) What is probability for recieved sequence being decoded correctly ?

(b) When $1$ million bits are sent over an information channel using sequences, what is expected value for incorrectly decoded sequences ?

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  • $\begingroup$ Do you always send five of the same bit? That is, if you want to transmit $0$ you always send $00000$? It seems implied in the question. Then you are just asked the chance that less than three of the bits get flipped to $1$. Look at the binomial distribution. $\endgroup$ Commented Sep 24, 2017 at 13:44
  • $\begingroup$ Yes this is the correct interpretation of this question $\endgroup$
    – Tuki
    Commented Sep 24, 2017 at 13:48
  • $\begingroup$ So can you compute the chance that all five bits are received correctly? How about four of them? $\endgroup$ Commented Sep 24, 2017 at 13:56
  • $\begingroup$ 10101 is interpreted as 1. So if we are sending 11111 I want to know what is the probability that our sequence will be interpreted as 1 on the receiving end. ( this could be for example we send 1111 it will be 10101 at the receiving end due loss in information channel and this would be interpreted still as 1. $\endgroup$
    – Tuki
    Commented Sep 24, 2017 at 13:57
  • $\begingroup$ Yes, you have to receive at least three bits properly to interpret the sequence correctly. Can you compute the chance of receiving $11111$ given that you sent $11111$? You have the chance of getting one bit right. $\endgroup$ Commented Sep 24, 2017 at 13:59

1 Answer 1

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Note that if $X$ is the number of bits that are transferred correctly, then $X$ follows a binomial distribution with parameters of selecting $5$ bits at a time and $0.9$ probability of a bit having been transferred correctly:

$$ \operatorname{P}(X=r)=\binom{5}{r}(0.9)^r(0.1)^{5-r} $$

For a sequence to be decoded correctly, only three of the five bits need to be transferred accurately. Thus the probability that the sequence is decoded correctly is

$$ \operatorname{P}(X\ge3)= \operatorname{P}(X=3)+ \operatorname{P}(X=4)+ \operatorname{P}(X=5) =\sum_{r=3}^5 \binom{5}{r}(0.9)^r(0.1)^{5-r}$$

I’ll leave that arithmetic to you.

The probability that a sequence is decoded incorrectly is $1- \operatorname{P}(X\ge3)$, which we just solved. Expected value for number of successes follows the form (number of trials)×(probability of success). Thus you can expect

$$\frac{10^6}{5} \left[ 1- \operatorname{P}(X\ge3)\right]$$

sequences to be decoded incorrectly since there are $\left. 10^6 \middle/ 5 \right.$ sequences transmitted.

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  • $\begingroup$ $E(x)=\frac{10^6}{5}∗(1−0.99144)=1712$ doesn't seem to give a correct answer? $E(x) = np$ is true for binomial distributions so in a sense this should work but there most me something I misunderstood? n = (number of trials) and p = (probability for success). However $P(X\ge 3)=0.99144$ is correct $\endgroup$
    – Tuki
    Commented Sep 24, 2017 at 21:15
  • $\begingroup$ @Tuki Could you elaborate on what you mean by that it “doesn’t seem to give a correct answer”? $\endgroup$ Commented Sep 24, 2017 at 21:26
  • $\begingroup$ Well this is for school and answers to these will be submitted to online form which then tells you if it is correct or not. 1712 is incorrect answer according to this system. $\endgroup$
    – Tuki
    Commented Sep 24, 2017 at 21:32
  • $\begingroup$ @Tuki That’s very weird, because (a) is the hard part and (b) is the easy part. Perhaps this “system” intended to ask a slightly different question than what I interpreted. $\endgroup$ Commented Sep 24, 2017 at 21:39
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    $\begingroup$ Correct answer to this question was $10^6 [1-P(X\ge 3)]$. There was issue with the assignment itself. So solution provided here is correct to how question is presented in the post. $\endgroup$
    – Tuki
    Commented Oct 5, 2017 at 15:39

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