Answer:
The expected or average no. of transmissions = 9.32
Explanation:
Chance of each frame to arrive undamaged = 80%
Total number of frames = 10
Probability of getting whole message to arrive undamaged = p = 0.80¹⁰
p = 0.1073
q = 1 - p = 1 - 0.1073 = 0.8927
where q is the probability of not getting whole message to arrive undamaged
The expected or average no. of transmissions = ∑
where k is from 0 to ∞
The above series can be reduced to p/(1-q)²
The expected or average no. of transmissions = p/(1-q)²
The expected or average no. of transmissions = 0.1073/(1-0.8927)²
Therefore, the expected or average no. of transmissions = 9.32