events

**z1llch** · July 15th 2008, 01:19 AM

events A and B are such that
p(A) = 1/2
p(A'|b) = 1/3
p(A $\cup$ B) = 3/5

determine P(B|A').
totally have no idea..
how can i define p(A $\cup$ B) = 3/5 btw?

**Simplicity** · July 15th 2008, 02:08 AM

Originally Posted by z1llch

events A and B are such that
p(A) = 1/2
p(A'|b) = 1/3
p(A $\cup$ B) = 3/5

determine P(B|A').
totally have no idea..
how can i define p(A $\cup$ B) = 3/5 btw?

General probability rules for your question...

$P(A') = 1 - P(A)$

$P(A \cup B) = P(A)+P(B)-P(A\cap B)$

$P(A|B) = \frac{P(A\cap B)}{P(B)}$

**mr fantastic** · July 15th 2008, 02:20 AM

Originally Posted by z1llch

events A and B are such that
p(A) = 1/2
p(A'|b) = 1/3
p(A $\cup$ B) = 3/5

determine P(B|A').
totally have no idea..
how can i define p(A $\cup$ B) = 3/5 btw?

I'd suggest you draw a Karnaugh table. From the given data:

$\begin{tabular}{l | c | c | c} & A & A$\, '$ & \\ \hline B & a & b & \\ \hline B$\, '$ & c & (1/2) - b & \\ \hline & 1/2 & 1/2 & 1 \\ \end{tabular}<br />$

where:

$a + b + c = \frac{3}{5}$ .... (1)

$\frac{b}{a+b} = \frac{1}{3} \Rightarrow a = 2b$ .... (2)

Substitute (2) into (1): $c = \frac{3}{5} - 3b$ .

Update the Karnaugh table:

$\begin{tabular}{l | c | c | c} & A & A$\, '$ & \\ \hline B & 2b & b & 3b \\ \hline B$\, '$ & (3/5) - 3b & (1/2) - b & (11/10) - 4b \\ \hline & 1/2 & 1/2 & 1 \\ \end{tabular}<br />$

Therefore $3b + \frac{11}{10} - 4b = 1 \Rightarrow b = \frac{1}{10}$ .

Update the Karnaugh table:

$\begin{tabular}{l | c | c | c} & A & A$\, '$ & \\ \hline B & 2/10 & 1/10 & 3/10 \\ \hline B$\, '$ & 3/10 & 4/10 & 7/10 \\ \hline & 1/2 & 1/2 & 1 \\ \end{tabular}<br />$

Now it's easy to get $\Pr(B | A') = \frac{\Pr(B \cap A')}{\Pr(A')} = \frac{1/10}{1/2} = \frac{1}{5}$ .

**z1llch** · July 15th 2008, 02:41 AM

that karnaugh table was great! i don't think it's in my syllabus though

learnt a new thing, thanks a lot

the question suggested venn diagram, but i sucked in that. can we 'possibly' just do it with general rules lol?

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