# Solution to Problem 3 Let's denote: - \( P(E|1) \): Probability of error when 1 is transmitted = .45 - \( P(E|) \): Probability of error when is transmitted = .25 - \( P(1) \): Probability that a 1 is sent = .35 - \( P() \): Probability that a is sent --- ## 1. Probability of Receiving the Bit Correctly When a 1 is Transmitted \[ P(\text{Correct}|1) = 1 - P(E|1) = 1 - .45 = \boxed{.55} \] --- ## 2. Probability that a is Transmitted Since only or 1 can be transmitted: \[ P() = 1 - P(1) = 1 - .35 = \boxed{.65} \] --- ## 3. Total Probability That Any Bit is Received in Error \[ P(E) = P(E|1)P(1) + P(E|)P() \] \[ = (.45)(.35) + (.25)(.65) = .1575 + .1625 = \boxed{.32} \] --- ## 4. Total Probability That Any Bit is Received Correctly \[ P(C) = 1 - P(E) = 1 - .32 = \boxed{.68} \] Alternatively, you could calculate directly: \[ P(C) = P(C|1)P(1) + P(C|)P() \] \[ P(C|1) = 1 - P(E|1) = .55 \] \[ P(C|) = 1 - P(E|) = .75 \] \[ P(C) = (.55)(.35) + (.75)(.65) = .1925 + .4875 = .68 \] --- ## 5. Probability that a was Transmitted Given a was Received We want \(P(|\text{received })\). Use Bayes’ theorem: \[ P(|\text{rec }) = \frac{P(\text{rec }|)P()}{P(\text{rec })} \] Where: - \(P(\text{rec }|) = 1 - P(E|) = .75\) - \(P(\text{rec }|1) = P(E|1) = .45\) - \(P(\text{rec }) = P(\text{rec }|)P() + P(\text{rec }|1)P(1)\) \[ = (.75)(.65) + (.45)(.35) = .4875 + .1575 = .645 \] So, \[ P(|\text{rec }) = \frac{.75 \times .65}{.645} = \frac{.4875}{.645} \approx \boxed{.756} \]