Markov chain, probability of n people being infected after t periods.

Question

Suppose there are n people holding hands in a line where the first and last person are only holding hands with a single other person. Suppose the first person in the line is infected and every period he coughs and with probability p, his neighbor catches the virus. However, every period there is also some probability q, that the virus vaccine is created and everybody is instantly cured (an absorption state that can be reached from anywhere).

Note that the probability that at least one person is infected at time 1 is 100% since patient 1 is the original infected person. I am interested in two things, what is the probability that after T periods, EXACTLY N people are infected and what is the probability that at LEAST N people are infected.

I think the probability that at least two people are infected after T periods is:

$$ \sum_{t=1}^{T}p(1-p-q)^{t-1}$$

So I think that the probability that at least 3 people are infected after T periods is

$$ \sum_{t=2}^{T}p^2(t-1)(1-p-q)^{t-2} $$

Answer 1

Let $p_k$ denote that the probability that $k$ people are affected at time $T$. Clearly then, the probability that least $n$ people are affected is given by $\sum_{i = k}^n p_i$, so knowing $p_i$ is sufficient for both questions.

Let's fix some $1 \le k \le n$ and compute $p_k$. We need to propagate the virus exactly $k-1$ times over $T$ periods, so

• there are $\binom{T}{k-1}$ ways to choose which steps will be the ones propagating
• propagating steps must carry the extra weight of $p$ each, i.e. $p^{k-1}$ total weight
• all steps must carry the weight of $1-q$ to make sure there is no vaccine, i.e. a total weight of $(1-q)^T$

Thus the total amount is $$p_k = \binom{T}{k-1} p^{k-1} (1-q)^T.$$

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All that is assuming the event of creating a vaccine is a separate event from the disease propagation, i.e. at every step,

• vaccine is invented with probability $q$
• disease propagates with probability $(1-q)p$
• disease does not propagate with probability $(1-q)(1-p)$

If instead, the propagation probability is $p$ and not propagation is $1-p-q$, the numbers would be different.