Question 1069288: Consider a transmitter that is sending messages over a computer network. We define the random variable X to be the travel time of a message and Y to be the length of a message. Assume Y can take two possible values y = 10^2 bytes with probability 5/6 and y = 10^4 bytes with probability 1/6. The travel time of a message depends on both its length and random factors such as congestion in the network. The travel time is defined as 10^−4Y seconds with probability 1/2, 10^−3Y seconds with probability 1/3, and 10^−2Y seconds with probability 1/6.
Find the probability mass function of X and E(X).
I tried solving it but I was stuck at finding the e(x), below is my attempt. any hints or help would be appreciated.
let x = Travel time to a message
y = Length of message
p(Y) = { 5/6 if y= 10^2, 1/6 if y = 10^4}
p(X|Y) = { 1/2 if 10^-4 * Y, 1/3 if 10^-3 * Y , 1/6 if 10^-2* Y}
e(x) = sum xi * pi
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Start with a tree diagram
X ========p(x)=============X*p(x)
10^-2======5/12=15/36=======.004167; I get 10^-2 from 10^-4*10^2, probabilities 1/2 and 5/6
10^-1======5/18=10/36=======.027778; probabilities 1/3 and 5/6
1=========8/36============.222222; two-5/6*1/6 and 1/6 and 1/2
10========1/18=2/36========.555556
100=======1/36============2.777778
E(X)=3.5875 seconds.
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