[Math] Calculating Probability of a digital signal in a noisy channel!!

probability

A digital signal “1” or “0” is transmitted through a noisy channel, the received data may be different from the signal sent out. Suppose the transmitter sends out “0” with probability 0.6, and “1” with probability 0.4. When “0” is transmitted, the receiver receives “0” with probability 0.9, and “1” with probability 0.1. When “1” is transmitted, the receiver receives “0” with probability 0.2, and “1” with probability 0.8. 

 (1)  Find the probability that data “0” is received. 
 (2)  Find the probability of error (i.e. the probability that “1” is received when “0” was transmitted or “0” is received when “1” was transmitted).

I am trying to use conditional probability. For the first question
we will have two cases where , 0 is received when 0 is transmitted and 0 is received when 1 is transmitted. But I am going wrong.

Can anyone please explain and guide me through ?

Best Answer

1) You have two possibilities for receiving data "0".

  1. possibility is to get it when the transmitter sends 0. When transmitter sends 0, there is a 0.9 probability that "0" data will be received, and probability of sending a data "0" is 0.6 --- so for this possibility the probability is 0.9*0.6=0.54
  2. possibility is to get it when the transmitter sends 1. When transmitter sends 1, there is a 0.2 probability that "0" data will be received, and probability of sending a data "1" is 0.2 --- so for this possibility the probability is 0.4*0.2=0.08

This is a disjunction between this 2 possibilities (we only need at least one true) so we will add these 2 probabilities --- 0.54+0.08 = 0.62

2) Again, we have 2 possibilities in here :

A) The transmitter sends a "0" data and receiver receives an "1" data : probability for this is : 0.6 * 0.1 = 0.06

B) The transmitter sends a "1" data and receiver receives an "0" data: probability for this is : 0.4 * 0.2 = 0.08

This is a disjunction between this 2 possibilities (we only need at least one true) so we will add these 2 probabilities --- 0.06+0.08 = 0.14

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