1. ## Proabability problem

Hi all, Ive got an exam coming up in just over a week which will contain a question like this one.

From the attached Venn diagram work out

1) Determine P(A) and P(B)
2) Determine P(B|A) and P(A|B)
3) Determine P(¬B|A) and compare it with P(B|A)
4) Determine P(B|¬A)
5) Determine P(¬B|¬A) and compare it with P(B|¬A)
6) Determine P(¬A|¬B)

cheers.................

2. 3) $P\left( {\neg B|A} \right) = \frac{{13}}{{824}}$.

4) $P\left( {B|\neg A} \right) = \frac{{124}}{{176}}$.

Can you finish?

3. Originally Posted by marciano1976
Hi all, Ive got an exam coming up in just over a week which will contain a question like this one.

From the attached Venn diagram work out

1) Determine P(A) and P(B)
2) Determine P(B|A) and P(A|B)
3) Determine P(¬B|A) and compare it with P(B|A)
4) Determine P(B|¬A)
5) Determine P(¬B|¬A) and compare it with P(B|¬A)
6) Determine P(¬A|¬B)

cheers.................
3) P(¬B|A) is the probability that B will occur given that A has occured. You know P(A). Now you just need the probability that B won't occur and A will, which is 13/Total. So you just do P(¬B and A)/P(A) = 13/824. Notice, this is 1-P(B|A)

4) This is the probability that B will occur given that A hasn't occured. You know P(¬A)=1-P(A). So, you just need P(B and ¬A), which is 124/Total. So it's just P(B and ¬A)/P(¬A). Notice that this will be [P(B)-P(B|A)P(A)]/P(¬A) by the law of total probability

You should get the idea now.

4. ## parts 5 and 6

Hi, Thanx for your replies to my problem.

For part 5 i get

52/76 = 0.2954

For part 6 i get

52/65 = 0.8

i don`t think these are near the mark however.....Help!!!!

5. This is what I got.......

Let $n(\varepsilon)$ denote the total number of elements in the universal set.

Question 3

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

From the diagram we see that $Pr(B' \cap A)=\frac{13}{n(\varepsilon)}$

$\Rightarrow Pr(B'|A)=\frac{(\frac{13}{n(\varepsilon)})}{(\frac {811+13}{n(\varepsilon)})}=\frac{13}{824}$

Question 4

Similar method to 3

Question 5

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

$Pr(B' \cap A')=\frac{52}{n(\varepsilon)}$

$= \frac{(\frac{52}{n(\varepsilon)})}{(\frac{824}{n(\ varepsilon)})}=\frac{52}{824}=\frac{13}{206}$

Question 6

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

$\Rightarrow Pr(A'|B')=\frac{(\frac{52}{n(\varepsilon)})}{(\fra c{935}{n(\varepsilon)})}=\frac{52}{935}$

6. DivideBy0,
$P\left( {B|\neg A} \right)$