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	<title>Comments on: Statistics question I need help on?</title>
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		<title>By: nothereanymoreomgteh</title>
		<link>http://sportcarx.com/2010/03/statistics-question-i-need-help-on/comment-page-1/#comment-835</link>
		<dc:creator>nothereanymoreomgteh</dc:creator>
		<pubDate>Sun, 21 Mar 2010 11:30:58 +0000</pubDate>
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		<description>P(Red) = 0.38, then P(Not Red) = 1 - P(Red) = 0.62

P(Not Van) = 1 - P(V) = 1 - 0.24 = 0.76

For the second part, you might want to check your numbers since the total does NOT add up to 1.</description>
		<content:encoded><![CDATA[<p>P(Red) = 0.38, then P(Not Red) = 1 &#8211; P(Red) = 0.62</p>
<p>P(Not Van) = 1 &#8211; P(V) = 1 &#8211; 0.24 = 0.76</p>
<p>For the second part, you might want to check your numbers since the total does NOT add up to 1.</p>
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		<title>By: Karen</title>
		<link>http://sportcarx.com/2010/03/statistics-question-i-need-help-on/comment-page-1/#comment-834</link>
		<dc:creator>Karen</dc:creator>
		<pubDate>Sun, 21 Mar 2010 11:16:23 +0000</pubDate>
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		<description>First one is asking for the probability of &quot;failure&quot; so to speak so the answer is 1 - 0.38

Second one:

We want the probability of not renting a van, which is 1 - P(V) = 1 - .24</description>
		<content:encoded><![CDATA[<p>First one is asking for the probability of &#8220;failure&#8221; so to speak so the answer is 1 &#8211; 0.38</p>
<p>Second one:</p>
<p>We want the probability of not renting a van, which is 1 &#8211; P(V) = 1 &#8211; .24</p>
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