Computing Probability

suzychevy

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Mar 6, 2010
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I've been working on stats for 13 hours. My brain is fried, my butt is asleep. I know this is simple. This is the last question I have to do. Can someone please show me how to do it and what the right answer should be???? Thank you.

If P(A) = 0.4, P(B | A) = 0.35, P(A È B) = 0.69, then P(B) =
 
Hello, suzychevy!

You must have used the wrong code.
I'll take a guess at what you meant.


P(A)=0.4P(BA)=0.35P(AB)=0.69\displaystyle P(A) = 0.4\quad P(B|A) = 0.35 \quad P(A \cup B) = 0.69

\(\displaystyle \text{Find: }\:p(B)\)

Bayes’ Theorem:   P(BA)  =  P(AB)P(A)\displaystyle \text{Bayes' Theorem: }\;P(B|A) \;=\;\frac{P(A \cap B)}{P(A)}

So we have:   0.35  =  P(AB)0.4P(AB)=0.14\displaystyle \text{So we have: }\;0.35 \;=\;\frac{P(A \cap B)}{0.4} \quad\Rightarrow\quad P(A \cap B) \:=\:0.14


Formula:   P(AB)  =  P(A)+P(B)P(AB)\displaystyle \text{Formula: }\;P(A \cup B) \;=\;P(A) + P(B) - P(A \cap B)

So we have:     0.69=      0.4    +P(B)        0.14\displaystyle \text{So we have: }\;\;0.69 \quad=\;\;\;0.4 \;\;+ P(B) \;\;-\;\; 0.14


Therefore:   P(B)  =  0.690.4+0.14  =  0.43\displaystyle \text{Therefore: }\;P(B) \;=\;0.69 - 0.4 + 0.14 \;=\;\boxed{0.43}

 
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