Probability-Not sure what this problem is asking me.

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holdyourhorsiieees

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Sep 21, 2007
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I'm given a problem that states:
"Consider rolling a single die. Define the following events: A={1,2,3,4}, B={1,3,5}, C={2,4,6}."

Then I am aksed to find A u C and P(A u C), etc.

I'm not exactly sure what's going on here...What are the event sample spaces telling me?
 
\(\displaystyle \begin{array}{l}
P\left( {A \cup C} \right) = P(A) + P(C) - P\left( {A \cap C} \right) \\
P\left( {\left\{ {1,2,3,4,6} \right\}} \right) = P\left( {\left\{ {1,2,3,4} \right\}} \right) + P\left( {\left\{ {2,4,6} \right\}} \right) - P\left( {\left\{ {2,4} \right\}} \right) \\
\end{array}\)
 
pka said:
\(\displaystyle \begin{array}{l}
P\left( {A \cup C} \right) = P(A) + P(C) - P\left( {A \cap C} \right) \\
P\left( {\left\{ {1,2,3,4,6} \right\}} \right) = P\left( {\left\{ {1,2,3,4} \right\}} \right) + P\left( {\left\{ {2,4,6} \right\}} \right) - P\left( {\left\{ {2,4} \right\}} \right) \\
\end{array}\)
Click to expand...

The problem says to calculate A u C and also the P(A u C). What is the difference in calculating those two?

Also, I'm not seeing where the P{2,4,6} and P{2,4} came from.
 
\(\displaystyle \begin{array}{l}
P\left\{ {2,4,6} \right\} = \frac{3}{6}\quad ,\quad \mbox{probability of an even.} \\
P\left\{ {2,4} \right\} = \frac{2}{6}\quad ,\quad \mbox{probability of an even less than six.} \\
\end{array}\)
 
holdyourhorsiieees said:
The problem says to calculate A u C and also the P(A u C). What is the difference in calculating those two?
Click to expand...

A U C is an event. It is a subset of the sample space. P(A U C) is the probability of satisfying the event. i.e. \(\displaystyle \frac{|A \cup C|}{|S|}\).
 
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