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I'm having lots of trouble with this question.. don't even know where to start.

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 = 102 bytes with probability 5/6 and y = 104 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(^-4)Y seconds with probability 1/2, 10(^-3)Y seconds with probability 1/3, and 10(^−2)Y seconds with probability 1/6. Find the probability mass function of X and E(X).

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    Welcome to Math.SE! [Probability theory](http://en.wikipedia.org/wiki/Probability_theory) is about the measure-theoretic foundations of stochastics. The tag ([tag:probability-theory]) should be used for questions concerning this subject, not for questions about calculating a specific probability. Use ([tag:probability]) instead, see also [meta](http://meta.math.stackexchange.com/questions/1686/the-tags-probability-and-probability-theory). – 2017-02-26

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This is a simple application of conditional probability. $\mathbb{P}[X=x]=\sum_{y}\mathbb{P}[Y=y]\mathbb{P}[X=x|Y=y]$. As each $x$ can occur for only one $y$ these calculations are straight-forward, for example $X=10^{-4}(102)$ occurs with probability $\frac{5}{6}\frac{1}{2}=\frac{5}{12}$. Similarly $\mathbb{E}[X]=\mathbb{E}[\mathbb{E}[X|Y]]=\mathbb{E}[Y\left(10^{-4}\frac{1}{2}+10^{-3}\frac{1}{3}+10^{-2}\frac{1}{6}\right)]=102.33\left(10^{-4}\frac{1}{2}+10^{-3}\frac{1}{3}+10^{-2}\frac{1}{6}\right)$.