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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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  • $\begingroup$ I think you mean $P[G]=0.5, P[B]=0.4, P[W]=0.8$. You might consider the identity $(W^c \cap B^c \cap G^c) \cup (W^c \cap B^c \cap G) = (W^c \cap B^c)$. $\endgroup$ – Michael Feb 2 '17 at 6:30
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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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    $\begingroup$ Your first line does not seem justified. $\endgroup$ – Michael Feb 2 '17 at 6:34
  • $\begingroup$ Really? If W is independent of G and B, then isn't any function of W independent of any function of G and B? $\endgroup$ – Nathaniel Mayer Feb 2 '17 at 6:36
  • $\begingroup$ No, but, in this case it might "accidentally" be true. $\endgroup$ – Michael Feb 2 '17 at 6:37
  • $\begingroup$ @Michael, if $W$ is independent of $G$ and $W$ is independent of $B$, then why isnt it always true that $W$ is independent of $G \cap W$? $\endgroup$ – Aditya Kalra Feb 2 '17 at 6:38
  • $\begingroup$ Well, if you are making such a claim, how do you prove it? $\endgroup$ – Michael Feb 2 '17 at 6:39

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