# Probability qn 2

• May 19th 2013, 07:14 AM
Tutu
Probability qn 2
Hi I really need help with this too! I cannot see where I went wrong, really. Thank you so much in advance! Really need this!

! I've attached the question itself and my attempt to find P(D) to which I failed.

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• May 19th 2013, 10:47 AM
Plato
Re: Probability qn 2
\begin{align*} \mathcal{P}(C) &= \mathcal{P}(C\cap D)+ \mathcal{P}(C\cap D') \\\frac{9}{20}&=\frac{6}{13}\mathcal{P}(D)+\frac{3 }{7}(1-\mathcal{P}(D))\end{align*}

Solve for $\mathcal{P}(D).$
• May 19th 2013, 11:35 PM
Tutu
Re: Probability qn 2
Thank you so much!
I didn't know that P(C|D)+P(C|D') = P(C)!
If I expand it out eg. P(C|D)P(D) = P(CnD), it will be quite different from P(C|D')P(D') = P(CnD') so I dont think that's the proof..

Could you teach me how to prove P(C|D)+P(C|D') = P(C) using any method or even the venn diagram?
• May 20th 2013, 02:33 AM
Plato
Re: Probability qn 2
Quote:

Originally Posted by Tutu
I didn't know that P(C|D)+P(C|D') = P(C)!
If I expand it out eg. P(C|D)P(D) = P(CnD), it will be quite different from P(C|D')P(D') = P(CnD') so I dont think that's the proof..

That is not true. Nor did I claim it was.

But it is true that
$\mathcal{P}(C)=\mathcal{P}(C\cap D)+\mathcal{P}(C\cap D')=\mathcal{P}(C|D)\mathcal{P}(D)+\mathcal{P}(C|D ')\mathcal{P}(D')$.
• May 20th 2013, 08:07 AM
Tutu
Re: Probability qn 2
I understand with the help of venn diagrams! Thank you so much!