In a communication system each message is sent 4 times over an unreliable channel. Assume that individual transmissions are independent from each other. It is known that the probability to receive the message correctly at least once is 0.9984.

How many times on average would one need to retransmit a message before it is received correctly?

  • 1
    $\begingroup$ You can use independence to find the probability that any individual transmission is successful (Hint: what is the probability that all 4 fail?). Then use this probability to define a geometric random variable, whose expectation is your desired result. $\endgroup$
    – platty
    Aug 7 '17 at 15:53

You have $P(X\ge1|N=4)=0.9984$ where $X$ is the number of correct transmissions and $N$ is the number of total transmissions. So you can work out $P(X=0|N=4) = 1-P(X\ge1|N=4)$. But also you have $P(X=0|N=4) = q^{4}$ because the transmissions are independent, where $q$ is the probability of a single transmission being incorrect (hence the probability of one transmission being correct is $1-q$ as those are the only two possibilities).

From there, you have a geometric distribution, since you keep sending the message until it is received correctly. The expected number of trials for a geometric distribution is $1/p$, where $p$ is the probability of success.

Answer below for verification:

$q^{4} = 0.0016 \Rightarrow q=0.2 \Rightarrow p=0.8$ so you need $1/0.8=1.25$ transmissions on average to transmit a message correctly.


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