Math-Law-other-studies: Second $$\LaTeX$$ test: a lot more code.

Saturday, January 13, 2007

Second $$\LaTeX$$ test: a lot more code.

This should now show a graphic containing $LaTeX$ written homework for one problem from last term's introduction to probability. $Problem 1.2-6 (c)\\ \\ Find \begin{math}P(A' \bigcup B' ) \end{math} \\ \\ Again using basic properties and theorems of probability\\ \\ P(A') =1-P(A) = 1-0.4 =0.6 and P(B') = 1-P(B) = 1- 0.5 = 0.5\\ \\ Now we need to use this formula\\ \\ \begin{math}P(A' \bigcup B') = P(A') + P(B') - P( A' \bigcap B')\end{math} in this problem.\\ \\ But \begin{math} P (A \bigcup B) = 0.6\end{math} and \begin{math}P( (A \bigcup B)')\end{math} its complement is simply\\ \\ \begin{math} 1- P (A \bigcup B) = 1- 0.6 =0.4\end{math} \\ \\ again by basic probability theorems\\ \\ What is the set \begin{math} (A \bigcup B)'\end{math} actually? It is the area out side of both A and B. which is the same as an intersection of the complements of A and B or set \begin{math}( A' \bigcap B')\end{math}\\ \\ Therefore \begin{math}P( A' \bigcap B') = 0.4\end{math}\\ \\ Therefore substituting in our formula.\\ \\ \begin{math}P(A' \bigcup B') = 0.6 + 0.5 - 0.4\end{math} = 0.7\\ \\ And the set \begin{math}(A' \bigcup B') \end{math} is really the complement of \begin{math}( A \bigcap B)\end{math} which = 0.3 because outside of the intersection is outside of either A or B so all this space outside the intersection is in \begin{math}(A' \bigcup B') \end{math} . Thus we can check our work by checking that \begin{math} P(A' \bigcup B') \end{math} = 1- \begin{math}P( A \bigcap B)\end{math} = 1- 0.3 = 0.7 \\$

Math-Law-other-studies

Saturday, January 13, 2007

Second $$\LaTeX$$ test: a lot more code.

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