Asked by John

A binary communication system is used to send one of two messages:

(i) message A is sent with probability 2/3, and consists of an infinite sequence of zeroes,

(ii) message B is sent with probability 1/3, and consists of an infinite sequence of ones.

The ith received bit is “correct" (i.e., the same as the transmitted bit) with probability 3/4, and is “incorrect" (i.e., a transmitted 0 is received as a 1, and vice versa), with probability 1/4. We assume that conditioned on any specific message sent, the received bits, denoted by Y1,Y2,… are independent.

Note: Enter numerical answers; do not enter '!' or combinations.

1) Find P(Y1=0), the probability that the first bit received is 0.
2)Given that message A was transmitted, what is the probability that exactly 6 of the first 10 received bits are ones? (Answer with at least 3 decimal digits.)
3)Find the probability that the first and second received bits are the same.
4)Given that Y1,…,Y5 were all equal to 0, what is the probability that Y6 is also zero?
5)Find the mean of K, where K=min{i:Yi=1} is the index of the first bit that is 1.

Answers

Answered by jeremy
1) 7/12
3) .625
4) .749267
The other questions I don't know the answers to.
Answered by Milorad
2) 210*((1/4)^6)*(3/4)^4
Answered by ubn
please tell the answer of fifth part.
Answered by 123
5) 3.1111
Answered by RVE
2. 0.01622
Answered by cle
From where did yoy get 210 (in the second answer)?

And can please you write the passage for point 5?
Answered by Jove
5567.678
Answered by GT
5) If K=1 then the probability is (1/3)*(3/4)+(2/3)*(1/4)
If K=2 then (1/3)*(1/4)*(3/4)+(2/3)*(3/4)*(1/4)
If K=3 then (1/3)*(1/4)^2(3/4)+(2/3)*(3/4)^2*(1/4)
For K=n then (1/3)*(1/4)^(n-1)(3/4)+(2/3)*(3/4)^(n-1)*(1/4)
We are looking for the expected value of K thus the sum(n=1 to n=inf.)(n*P)
Doing the math we conclude 28/9
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