Show that

$\frac{P(H\mid E)}{P(G\mid E)}=\frac{P\left(H\right)}{P\left(G\right)}\frac{P(E\mid H)}{P(E\mid G)}$

Suppose that, before new evidence is observed, the hypothesis $H$is three times as likely to be true as is the hypothesis $G$.

If the new evidence is twice as likely when $G$is true as it is when $H$is true, which hypothesis is more likely after the evidence has been observed?

The possibility of a happening or outcome occurring supported by the occurrence of a preceding event or outcome is understood as a contingent probability.

The updated probability of the next or conditional, event is multiplied by the probability of the preceding, or conditional, event to get introduce contingent probability.

Prove:

$\frac{P(H\mid E)}{P(G\mid E)}=\frac{P\left(H\right)}{P\left(G\right)}\times \frac{P(E\mid H)}{P(E\mid G)}$

$P\left(B\right)$isn't capable $0$for occurrences $A$and$B$

$P(A\mid B)=\frac{P(A\cap B)}{P\left(B\right)}$

Begin with the LHS and apply the definition:

$\frac{P\left(H\right|E)}{P\left(G\right|E)}=\frac{{\displaystyle \frac{P(H\cap E)}{P\left(E\right)}}}{{\displaystyle \frac{P(G\cap E)}{P\left(E\right)}}}$

$=\frac{P(H\cap E)}{P(G\cap E)}$

and on the RHS, using the identical definition:

$\frac{P\left(G\right)}{P\left(H\right)}\times \frac{P\left(E\right|H)}{P\left(E\right|G)}=\frac{P\left(H\right)}{P\left(G\right)}\times \frac{{\displaystyle \frac{P(H\cap E)}{P\left(H\right)}}}{{\displaystyle \frac{P(G\cap E)}{P\left(G\right)}}}$

=$=\frac{P(H\cap E)}{P(G\cap E)}$

As a result, it's been established.

The following statements are converted into mathematical equations.

$P\left(H\right)=3P\left(G\right)$

$P\left(E\right|G)=2P(E\left|H\right)$

Using the formula,

$$\begin{gathered} \frac{P\left(H\right|E)}{P\left(G\right|E)}=\frac{P\left(H\right)}{P\left(G\right)}\times \frac{P\left(E\right|H)}{P\left(E\right|G)} \\ =\frac{P\left(H\right)}{P\left(G\right)}\times \frac{P\left(E\right|H)}{P\left(E\right|G)} \end{gathered}$$

Replace the given facts together with your own.

localid="1649658537441" $$\begin{gathered} \frac{P\left(H\right|E)}{P\left(G\right|E)}=\frac{3P\left(G\right)}{P\left(G\right)}\times \frac{P\left(E\right|H)}{2P\left(E\right|H)} \\ =3\times \frac{1}{2} \\ =1.5 \end{gathered}$$

multiplying by $P\left(G\right|E)$and equating the primary and last expressions:

$P\left(H\right|E)=1.5\times P(G\left|E\right)$$\Rightarrow P(H\mid E)>P(G\mid E)$