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Suppose a large # of students are surveyed about how they travel, $G - 0.5$, $B - 0.4$, $W = 0.8$

Given that: $W$ is independent of $G$ and $W$ is independent of $B$, but $B$ is mutually exclusive of $G$, what is the probability that a random student does none of them?

We want $P(\overline{W} \cap \overline{G} \cap \overline{B})$, but how can I split it?

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    I think you mean $P[G]=0.5, P[B]=0.4, P[W]=0.8$. You might consider the identity $(W^c \cap B^c \cap G^c) \cup (W^c \cap B^c \cap G) = (W^c \cap B^c)$. – 2017-02-02

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My previous answer was flawed. It is not generally true that if $W$ is independent of $G$ and $B$, then it is independent of $G \cap B$.

It is true that independence of events is preserved under taking complements. Combining that with some use of the law of total probability gives a pretty quick computation, which happens to give the same answer I had before. \begin{align} P(W^c \cap B^c \cap G^c) &= P(W^c \cap B^c) - P(W^c \cap B^c \cap G) \\ &= P(W^c \cap B^c) - [P(W^c \cap G) - P(W^c \cap B \cap G)] \\ &= (0.2)(0.6) - (0.2)(0.5) + 0 \\ &= \boxed{0.02} \end{align}

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    Your first line does not seem justified. – 2017-02-02
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    Really? If W is independent of G and B, then isn't any function of W independent of any function of G and B? – 2017-02-02
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    No, but, in this case it might "accidentally" be true. – 2017-02-02
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    @Michael, if $W$ is independent of $G$ and $W$ is independent of $B$, then why isnt it always true that $W$ is independent of $G \cap W$? – 2017-02-02
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    Well, if you are making such a claim, how do you prove it? – 2017-02-02
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    A standard example is this: Let $X$ and $Y$ be independent and equally likely to be either $-1$ or $1$. Define $Z=XY$. Then every pair of random variables is independent, but they are not mutually independent. If I tell you that $X=1$ and $Y=1$, then you know $Z=1$ (even though the events $X=1$ and $Z=1$ are independent, as are the events $Y=1$ and $Z=1$). – 2017-02-02
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    You're absolutely right, that was totally false. Fixing my answer now – 2017-02-02
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    Also, sorry for changing notation. I got really tired of typing \overline{}. – 2017-02-02
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    How is $W^c \cap B^c \cap G^c = W^c \cap B^c - W^c \cap B^c \cap G)$? – 2017-02-02