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If probability of frame reaching safely is 0.1 then mean number of transmissions of a frame to make it success is _____.
 

what is wrong with the approach given below

‘random variable x is the number of transmission’

X 1 2 3 4…       so on
P(X) 0.1 0.9*0.1 0.9*0.9*0.1 0.9*0.9*0.9*0.1

mean or expectation value of random variable E[X]= $\sum X*P(X)$

which comes out to be 100.

but  given the answer is 10 using 1/p.

(what is wrong in above aproach?)

in Computer Networks by Active (1.7k points)
edited by | 87 views
+1

E[X]= ∑ ( XP(X) )

 = 1 * (0.1) * (0.9)$^0$ + 2 * (0.1) * (0.9)$^1$ + 3 * (0.1) * (0.9)$^2$ + 4 * (0.1) * (0.9)$^3$ .......... upto infinity

E(x) = (0.1) { 1 * (0.9)$^0$ + 2 * (0.9)$^1$ + 3 * (0.9)$^2$ + 4 * (0.9)$^3$ .......... upto infinity }

let S = 1 * (0.9)$^0$ + 2 * (0.9)$^1$ + 3 * (0.9)$^2$ + 4 * (0.9)$^3$ .......... upto infinity

let  k = 0.9 ===> (1-k) = 0.1, but note that k < 1

S = 1 * (k)$^0$ + 2 * (k)$^1$ + 3 * (k)$^2$ + 4 * (k)$^3$ .......... upto infinity ----------------- (1)

==> kS = k * {1 * (k)$^0$ + 2 * (k)$^1$ + 3 * (k)$^2$ + 4 * (k)$^3$ .......... upto infinity}

kS = 1 * (k)$^1$ + 2 * (k)$^2$ + 3 * (k)$^3$ + 4 * (k)$^4$ .......... upto infinity ----------------- (2)

Perform (1) - (2)

S - kS = 1 * (k)$^0$ + 1 * (k)$^1$ + 1 * (k)$^2$ + 1 * (k)$^3$ .......... upto infinity

S(1-k) = (k)$^0$ + (k)$^1$ + (k)$^2$ + (k)$^3$ + (k)$^4$ .......... upto infinity

S(1-k) = $\frac{1}{1-k}$  ------------- ( apply infinite sum of GP series where r < 1 )

S = $\frac{1}{(1-k)(1-k)}$

E(X) = 0.1 * S = $\frac{1}{(1-k)(1-k)}$ = 0.1 * $\frac{1}{(0.1)(0.1)}$ = $\frac{1}{(0.1)}$ = 10

0
calculation mistake from my side .thanks for answer

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