# Thread: Two events and a simple question

1. ## Two events and a simple question

what the question states :
"Events A and B are such that:
p(A) = 5/12
p(A|notB)=7/12
P(A intersection B ) =1/8 "

Would I be correct if I assume that Not B refers to A ? , so then i can draw a probability tree.

2. ## Re: Two events and a simple question

Originally Posted by terence
what the question states :
"Events A and B are such that:
p(A) = 5/12
p(A|notB)=7/12
P(A intersection B ) =1/8 "

Would I be correct if I assume that Not B refers to A ? , so then i can draw a probability tree.
no you cant assume that, not B not always just A, if A union B is the whole space then not B is in A

3. ## Re: Two events and a simple question

Most likely "notB" means "event B did not occur"

I would also advise you to draw a Venn diagram for this question.

4. ## Re: Two events and a simple question

Originally Posted by terence
what the question states :
"Events A and B are such that:
p(A) = 5/12
p(A|notB)=7/12
P(A intersection B ) =1/8 "
Would I be correct if I assume that Not B refers to A ? , so then i can draw a probability tree.
$\displaystyle P(A|B^c)=\frac{P(A\cap B^c)}{P(B^c)}$.
Now $\displaystyle P(B^c)=1-P(B)$.
Thus $\displaystyle P(A\cap B^c)=P(A|B^c)P(B^c)$

5. ## Re: Two events and a simple question

Originally Posted by Plato
$\displaystyle P(A|B^c)=\frac{P(A\cap B^c)}{P(B^c)}$.
Now $\displaystyle P(B^c)=1-P(B)$.
Thus $\displaystyle P(A\cap B^c)=P(A|B^c)P(B^c)$
The question then asks for :

i) P(B)
ii) P(A|B)
iii) P(B|A)
iv) P(A U B)

P(B|A) is easy since P(A intersection B) / P(A) = 0.3
Had i known P(B) iv) would have been easy as well.

How can P(B) be found.. ?
Well I know this:
\displaystyle \begin{align*}P(B) &= P(A\cap B) &+ P(notA\cap B)\end{align*}
where
\displaystyle \begin{align*}P(notA\cap B) &=P(A).P(B|notA)\end{align*}

6. ## Re: Two events and a simple question

Originally Posted by terence
The question then asks for :
i) P(B)
ii) P(A|B)
iii) P(B|A)
iv) P(A U B)
How can P(B) be found.. ?
Well I know this:
\displaystyle \begin{align*}P(B) &= P(A\cap B) &+ P(notA\cap B)\end{align*}
where
\displaystyle \begin{align*}P(notA\cap B) &=P(A).P(B|notA)\end{align*}
From the given you can find $\displaystyle P(A\cap B^c).$
But $\displaystyle P(A\cap B^c)=P(A|B^c)P(B^c)$ which is given.
From which you can find $\displaystyle P(B^c)$. From which you can find $\displaystyle P(B)$.

7. ## Re: Two events and a simple question

From the given you can find $\displaystyle P(A\cap B^C)$
how can you find this ?

the nearest i can get is that
$\displaystyle P(A\cap B^C) &= 7/12 * P(B)$

8. ## Re: Two events and a simple question

Originally Posted by terence
how can you find this ?
$\displaystyle P(A\cap B^c)=P(A)-P(A\cap B)$

9. ## Re: Two events and a simple question

Oh how dumb of me -.-

I should have reasoned that out. Largest problem with probability=> reasoning simple things out.

Thanks @Plato