Find P(B) and P(A/B) by considering the given probability question

[chwala]

Homework Statement

Find $P(B)$ and $P(A/B)$ by considering the given problem below(attached)

Relevant Equations

conditional probability

i managed to find the values...i am seeking alternative approach to the problem. See my attempt below.

 

[chwala]

 

[FactChecker]

You need to be more careful about each step. The first equality of your attempt is wrong. The Venn Diagram only has B' shaded, so which part is A$\cap$B'? It would help if you would shade A in one direction and B' in another direction and clearly show what A$\cap$B' is. The equality from the Venn diagram looks wrong. There may be more mistakes. You should look carefully at each step more carefully before proceeding.

 

[chwala]

$A$ intersection $B^{'}$ is the shaded region...it follows that $B$ is the unshaded...

 

[chwala]

Let me check on this after my lunch .

 

[FactChecker]

$A$ intersection $B^{'}$ is the shaded region...it follows that $B$ is the unshaded...

No, it is not. Much of that shaded area is not in A. And that would not be how you decide what B is anyway. You are correct that B is all of the inside of the B circle, but that is always true no matter what the rest of the Venn diagram looks like.

 

[chwala]

Ok give me a moment, let me check again...

 

[chwala]

My approach was wrong, the correct way is as follows;
$P(B/A)= \frac {P(B). P(A/B)}{P(A)}$
$0.8= \frac {P(B). P(A/B)}{0.75}$
$0.6=P(B). P(A/B)$

Also,
$P(B/A^{'})= \frac {P(B). P(A^{'}/B)}{P(A^{'})}$
$0.6= \frac {P(B). P(A^{'}/B)}{0.25}$
$0.15=P(B). P(A^{'}/B)$

we know that,
$P(A/B)+P(A^{'}/B)=1$
$\frac {0.6}{P(B)}$+$\frac {0.15}{P(B)}$=$1$
on cross multiplication by $P(B)$,
$0.6+0.15=P(B)$
$0.75=P(B)$...with this found the other part is easy...
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