Question

The transmitter transmits either an infinite sequence of 0s with a probability 2/3 or 1s with a probability 1/3. Each symbol, regardless of the others and the transmitted sequence is identified by the receiving device with an error with a probability 0.25. i) Given that the first 5 identified symbols are 0s, find the probability P (000000 | 00000) that the sixth received symbol is also zero. b) Find the average value of a random variable equal to the number of the first 1 written by the receiving device (for example, we received 00001...., our RV takes value 5).

Answer 1

(i) We are given here that:
P(0 infinite sequence) = 2/3,
P( 1 infinite sequency) = 1/3

Also, we are given here that:
P( error) = 0.25, therefore, P(no error) = 1 - 0.25 = 0.75

Therefore, Using law of total probability, we have here:
P( 00000 identified ) = [P(0 infinite sequence)P(no error)]⁵ + P( 1 infinite sequency)[P(error)]⁵

P( 00000 identified ) = (2/3)*0.75⁵ + (1/3)*0.25⁵ = 0.158529

Now using Bayes theorem, we have here:
P(0 infinite sequence | 00000) = [P(0 infinite sequence)P(no error)]⁵ / P( 00000 identified )
= (2/3)*0.75⁵ / 0.158529
= 0.9979

Therefore,
P(1 infinite sequence | 00000) = 1 - 0.9979 = 0.0021

Now, the required probability is computed as:
P( 000000 | 00000) = P(0 infinite sequence | 00000)*P(no error) + P(1 infinite sequence | 00000)*P(error)

= 0.9979*0.75 + 0.0021*0.25

= 0.7489

Therefore 0.7489 is the required probability here.

b) The expected value or the average value for the number on which the first 1 is written is computed here as the expected value of the geometric random variable computed as:

= 1 / p where p is the probability of getting a 1

= 1 / (1/3)

= 3

Therefore 3 is the required expected value here.