Respuesta :
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 = ∑ [tex]kp(1-p)^{k-1}[/tex]
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
The number of times that the message must be sent on average to get the entire thing through is; 10 times
The question did not tell us anything about datalink protocols.
Now, let us assume that there will be a successful transmission of the whole fragmented message with the single condition that all the frames arrive intact.
Thus;
P(successful transmission of the message) = P(all the frames arrive intact); p = 0.8¹⁰
Now, probability of unsuccessful transmission of the message;
q = 1 - p
Now, the Mean value of number of tries is simply the Expected value of k.
The formula for that is;
Σ[k][P(k)]
where k is from 1 to ∞
The binomial expansion of that formula gives us;
1p + 2qp + 3q²p + 4q³p .....
This can be reduced to;
p(1 + 2q + 3q² + 4q³ .....)
This can be further reduced to;
p/(1 - q)²
Recall that; q = 1 - p
Thus;
p/(1 - q²) = p/(1 - (1 - p))²
⇒ 1/p
Put 0.8¹⁰ for p to get;
1/0.8¹⁰ = 9.31
Thus, mean number of tries will be approximately 10.
Read more about expected value at; https://brainly.com/question/25926316
