This question extends a previous question I asked with respect to the markov chain model. I post another question here since I'm trying to follow the Stack Exchange's guidelines.
So, based on the same context as before, let $X_i$ be the number of messages left in Alice's mailbox after she checks and answers them on day $i$. For instance, after Alice checks on day 2, $X_2 = 1$ implies that 1 message is remaining in the mailbox.
By multiplying the transition probabilities along the path, the computations that follow are all based on $X_1=0$. For reference, these are all the transition probabilities:
$$\begin{align} p_{00}&=q_0+q_1(p_1+p_2)+q_2p_2=0.7\\ p_{01}&=q_1p_0+q_2p_1=0.22\\ p_{02}&=q_2p_0=0.08\\ \\ p_{10}&=q_0(p_1+p_2)+(q_1+q_2)p_2=0.53\\ p_{11}&=q_0p_0+(q_1+q_2)p_1=0.29\\ p_{12}&=(q_1+q_2)p_0=0.18\\ \\ p_{20}&=p_2=0.5\\ p_{21}&=p_1=0.3\\ p_{22}&=p_0=0.2\ . \end{align}$$
Firstly, to find the probability that $P(X_2=1,X_3=2,X_4=0 \space | \space X_1=0)$:
$$=(0.22)(0.18)(0.5)=0.0198$$
Similarly, given that $X_1=0$, the probability that after checking the mailbox on each of the next three days, there will never be more than two messages left, that is $X_i≠2$ for all $i=2,3,4$:
$$P(X_4=2,X_3=2,X_2=2 \space │ \space X_1=0)=(0.08)(0.2)^2=0.0032$$
The previous probability is for $X_i=2$, hence for $X_i≠2$ this is what I've got:
$$1-0.0032=0.9968$$
Lastly, by the same condition, given that $X_1=0$, the probability that $X_4=0$ is:
$$(X_4=0 \space │ \space X_1=0)=(0.7)^3=0.343$$
I was having a hard time trying to figure out conditional probabilities for path calculation.
Could anyone please tell me if my calculations are legit?
Perhaps someone could point out if there is a faster (or more efficient) way to calculate the desired probabilities.
Thanks
Your calculation of $\ P\big(X_2=1,X_3=2,X_4=0\,\big|\,X_1=0\big)\ $ is correct, and there is no simpler way of determining this quantity.
Your calculation of $\ P\big(X_4=0\,\big|\,X_1=0\big)\ $, however, isn't correct. What you have calculated is $\ p_{00}^3=P\big(X_2=0,X_3=0,X_4=0\,\big|\,X_1=0\big)\ $. But there are many other ways in which $\ X_1\ $ and $\ X_4\ $ could both be $\ 0\ $ without $\ X_2\ $ and $\ X_3\ $ also both being $\ 0\ $. You've already calculated the probability of one of them above (where $\ X_2=1\ $ and $\ X_3=2\ $). To obtain the correct value of $\ P\big(X_4=0\,\big|\,X_1=0\big)\ $ you need to sum $\ P\big(X_1=a,X_2=b,X_4=0\,\big|\,X_1=0\big)\ $ over all possible values $\ a\ $ and $\ b\ $ that $\ X_1\ $ and $\ X_2\ $ might have: \begin{align} P\big(X_4=0\,\big|\,X_1=0\big)&=\sum_{a=0}^2\sum_{b=0}^2P\big(X_2=a,X_3=b,X_4=0\,\big|\,X_1=0\big)\\ &=\sum_{a=0}^2\sum_{b=0}^2p_{0a}p_{ab}p_{b0}\\ &=\pmatrix{0.7&0.22&0.08}\pmatrix{0.7&0.22&0.08\\ 0.53&0.29&0.18\\ 0.5&0.3&0.2}\pmatrix{0.7\\0.53\\0.5}\\ &=0.636574\ . \end{align} In this case, treating the double sum as a matrix product doesn't save you any work, but if you're familiar with matrix multiplication, it does help you keep track of the computation. More generally, however, the conditional probability $\ P\big(X_t=b\,\big|\,X_1=a\big)\ $ for a finite-state Markov chain and $\ t\ge3\ $ is given by $$ P\big(X_t=b\,\big|\,X_1=a\big)=r_a\Pi^{t-3}c_b\ $$ where $\ \Pi\ $ is the transition matrix of the chain, $\ r_a\ $ is its $\ a^\text{th}\ $ row, and $\ c_b\ $ its $\ b^\text{th}\ $ column. When $\ t>4\ $, computing the matrix product is much less work than summing over all $\ n^{\,t-2}\ $ possible sequences of states ($\ n\ $ being the number of states—which is $3$ in your example) for $\ X_2,$$\,X_3,$$\,\dots,$$\,X_{t-1}\ $.
Your description of your remaining problem, which is
is not consistent. The event that $\ X_i\ne2\ $ for all $\ i=2,3,4\ $ is not the the same as the event that there will never be more than two messages left on the next three days, which, according to your description of the problem, is certain to be the case. The event that $\ X_i\ne2\ $ for all $\ i=2,3,4\ $ is the same as the event that there will never be more than one message left on the next three days. Assuming this is the event whose conditional probability you wish to calculate, then you haven't calculated it correctly.
The first conditional probability you get in your calculation is that of the event $\ \big\{X_2=X_3=X_4=2\big\}\ $. Then you subtract this from $\ 1\ $, which will give you the conditional probability of this event's complement. But the complement of this event is $\ \big\{X_2\ne2\ or\ X_3\ne2\ or\ X_4\ne2\big\}\ $, not $\ \big\{X_2\ne2\ and\ X_3\ne2\ and\ X_4\ne2\big\}\ $, which is the event whose probability you're apparently trying to calculate. To calculate the conditional probability of this latter event you have to sum the conditional probabilities of all the events $$\big\{X_2=i\ and\ X_3=j\ and\ X_4=j\big\}$$ over all possible values of $\ i\ne2\ $,$\ j\ne2\ $ and $\ k\ne2\ $: \begin{align} P\big(X_2\ne2,X_3\ne2,X_4\ne2\,\big|\,X_1=0\big)&=\sum_{i=0}^1\sum_{j=0}^1\sum_{k=0}^1P\big(X_2=i,X_3=j,X_4=j\,\big|\,X_1=0\big)\\ &=\sum_{i=0}^1\sum_{j=0}^1\sum_{k=0}^1p_{0i}p_{ij}p_{jk}\\ &=\pmatrix{0.7&0.22}\pmatrix{0.7&0.22\\ 0.53&0.29}^2\pmatrix{1\\1}\\ &=0.63469042\ . \end{align} This time using the matrix multiplication method can save you quite a bit of work over computing all eight probabilities $\ p_{0i}p_{ij}p_{jk}\ $ for all $\ i,j,k\ $ equal to $\ 0\ $ or $\ 1\ $ and then adding them.