Probability Question

[rsingh628]

Hi all, I have a probability question that I am working through and wanted to check to make sure I'm on the right track and get some guidance if I'm wrong. I've attached my scratch work, since it is a bit long. Thank you for the help.

Problem:


Approach attached:

 

[Steven G]

I guess we are assuming no ties??

P(win | won the last game)=.8
P(win | lost the last game)=.3
P(speech | lost)=.7
P(speech | won)=.2
P(speech) = P(speech and won) + P(speech and Lost)
Take it from here

 

[Steven G]

Where did this problem come from? It has a couple of typos that make me suspicious of the author having written the problem correctly.
I too am not getting an answer.

 

[BigBeachBanana]

Consider a Markov Chain with 2 states, $\Omega=\{0,1\}$. The winning/losing probabilities are given by the transition matrix $P:$

$$P=\left[\begin{matrix} 0.8 & 0.2\\ 0.3 & 0.7 \end{matrix}\right]$$
We are interested in knowing the fraction of the time the Markov Chain spent in each state (winning/losing) as n grows larger i.e. the limiting distribution $\lim_{n \to \infty} \pi^{(n)}$, where $\pi^{(0)}= [P(X_0=0)\quad P(X_0=1)]$.

$$\lim_{n \to \infty} P^n=\frac{1}{0.2+0.3} \left [\begin{matrix} 0.2 & 0.3\\ 0.2 & 0.3 \end{matrix} \right]$$
Then we have:
$$\lim_{n \to \infty} \pi^{(n)}= \lim_{n \to \infty} \pi^{(0)}P^{n}= \frac{}{}[P(X_0=0)\quad P(X_0=1)]\cdot \frac{1}{0.5} \left [\begin{matrix} 0.2 & 0.3\\ 0.2 & 0.3 \end{matrix} \right]= [0.6\quad 0.4]$$
Thus the probability of winning is $P(W) = 0.6$ and the probability of losing is $P(L) = 0.4$.
Additionally, we are given $P(S|W)=0.2, P(S|L)=0.7$
Therefore,
$$P(S) = P(S \cap W) + P(S\cap L)=P(S|W)P(W) + P(S|L)P(L)=(0.2)(0.6)+(0.7)(0.4)=\boxed{0.4}$$