Conditional probability in Markov chain

Question

Q: A simplified model for the spread of a contagion in a small population of size $4$ is as follows. At each discrete time unit, two individuals in the population are chosen uniformly at random to meet. If one of these persons is healthy and the other has the contagion, then with probability $\frac{1}{4}$ the healthy person becomes sick. Otherwise the system stays the same.

Assume now that the process begins with exactly one person infected. Given that not everyone is infected after $3$ steps of the process, what is the chance exactly one person is infected?

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A: I understand I am looking for $P(X_3=3|X_3>0,X_0=3)$. In the answer sheet though, it says the above is equal to the following $$\frac{P(X_3 =3 |X_0 =3)}{P(X_3 >0|X_0 = 3)}$$

Why are these equivalent? I know the the conditional proberty P(A|B)=P(A and B)/P(B) and the multiplication rule $P(A B)=P(A|B)\cdot P(B)$ but I can't seem to make the above work.

Thanks for your help!

Answer 1

I think you already have the answer.

Take $A$ as $(X_3=3|X_0=3)$ and $B$ as $(X_3>0|X_0=3)$ and note that in this case $P(A \cap B) = P(A)$ because $A$ implies $B$.

And you get $P(A|B) = P(A)/P(B)$

Answer 2

Think of

$A : \text{the event } X_3 = 3$;

$B: \text{the event } X_0 = 3 \wedge X_3>0$.

Then, we can write your equation: $P(X_3=3|X_3>0,X_0=3) = P(A|B). $

$$P(A|B) = \frac{P(AB)}{P(B)} = \frac{P(X_3=3,X_3>0,X_0=3)}{P(X_3>0,X_0=3)} =\frac{P(X_3=3,X_0=3)}{P(X_3>0,X_0=3)}$$

Since we know that $X_0 $ happened before $X_3$, $P(X_3=3,X_0=3)$ is equivalent to $P(X_3=3|X_0=3)$ and the same for $P(X_3>0,X_0=3) = P(X_3>0|X_0=3)$.

Hope this makes sense!