Question

Consider a transmitter that is sending messages over a computer network. Let us define the following two random variables: X : the travel time of a given message, Y : the length of the given message. We know the PMF of the travel time of a message that has a given length, and we know the PMF of the message length. We assume that the length of a message can take two possible values: y=102 bytes with probability 5/6, and y=104 bytes with probability 1/6, so that pY(y)={5/6,1/6, if y=102 if y=104 We assume that the travel time X of the message depends on its length Y and the congestion in the network at the time of transmission. In particular, the travel time is 10−4Y seconds with probability 1/2,10−3Y seconds with probability 1/3, and 10−2Y seconds with probability 1/6. Thus, we have pX∣Y(x∣102)=⎩⎨⎧1/2,1/3,1/6, if x=10−2, if x=10−1, if x=1,pX∣Y(x∣104)=⎩⎨⎧1/2,1/3,1/6, if x=1 if x=10 if x=100 Compute P(X=100) 1/18 1/36 1/24

Answer 1

To find P(X=100), multiply the conditional probability P(X=100 | Y=104) by the probability P(Y=104), which results in 1/36.

Step-by-step explanation:

To compute P(X=100), we first note that X=100 occurs only if Y=104 since the travel time is directly related to the message length. We can calculate P(X=100) as follows:

P(X=100) = P(X=100 | Y=104) × P(Y=104).

Given the PMF of X conditioned on Y, we know that P(X=100 | Y=104) = 1/6, as it is the probability that the message with 104 bytes will have a travel time of 100 seconds. Also, from the PMF of Y, we have P(Y=104) = 1/6. Thus:

P(X=100) = (1/6) × (1/6) = 1/36.

Hence, the probability that the travel time X equals 100 seconds is 1/36.