91Ó°ÊÓ

To prove:$\frac{P(Hâˆ\pounds E)}{P(Gâˆ\pounds E)}=\frac{P\left(H\right)}{P\left(G\right)}\frac{P(Eâˆ\pounds H)}{P(Eâˆ\pounds G)}$.

For events $A$and $B$such that $P\left(B\right)≠0$(so that the fraction is defined):

$P(Aâˆ\pounds B):=\frac{P(Aâˆ\copyright B)}{P\left(B\right)}$

Start from the left hand side, apply this definition:

$\frac{P(Hâˆ\pounds E)}{P(Gâˆ\pounds E)}=\frac{\frac{P(Hâˆ\copyright E)}{P\left(E\right)}}{\frac{P(Gâˆ\copyright E)}{P\left(E\right)}}=\frac{P(Hâˆ\copyright E)}{P(Gâˆ\copyright E)}$

And by using the same definition on the right hand side:

$\frac{P\left(H\right)}{P\left(G\right)}·\frac{P(Eâˆ\pounds H)}{P(Eâˆ\pounds G)}=\frac{P\left(H\right)}{P\left(G\right)}·\frac{\frac{P(Hâˆ\copyright E)}{P\left(H\right)}}{\frac{P\left(G\right)E)}{P\left(G\right)}}=\frac{P(Hâˆ\copyright E)}{P(Gâˆ\copyright E)}$

As both sides equal to the same expression the properties of equality prove the given statement.

Given statements are transferred into mathematical equations:$P\left(H\right)=3P\left(G\right)$

$P(Eâˆ\pounds G)=2P(Eâˆ\pounds H)$

Using the equation from part I):

$\frac{P(Hâˆ\pounds E)}{P(Gâˆ\pounds E)}=\frac{P\left(H\right)}{P\left(G\right)}·\frac{P(Eâˆ\pounds H)}{P(Eâˆ\pounds G)}=\frac{P\left(H\right)}{P\left(G\right)}·\frac{P(Eâˆ\pounds H)}{P(Eâˆ\pounds G)}$

Substitute the given facts:

Equating the first and the last expression, and multiplying by $P(Gâˆ\pounds E)$ :

To prove the identity use the definition of conditional probability

$P(Hâˆ\pounds E)>P(Gâˆ\pounds E)$ this follows from the given statements and the proven equation.