# Linear system with probabilities (algebra)

I have a small problem that delays my project , it seems I am stuck here loads of time. It is probably very easy but I cant see it right now , I am very anxious about this, please take a look: \begin{align*} P_0 &= \frac45P_1 + \frac34P_1\\ P_1 &= \frac15P_0 + \frac14P_1 \end{align*} also I know this: $P_1+P_0=1$

The solution is $P_0= 15/19$ and $P_1= 4/19$.

I have a very similar problem and I don't know how to continue from this point on and delays my project!! How am i supposed to solve this ? For more probabilities like $P_1, P_2, P_3$ what can we do ? Is there some general algorithm for this with steps ?

Thanks a lot!

-
Is it $(4/5)Po$ or $\frac{4}{5 PO}$? Same for the others. Regards – Amzoti Jan 17 '13 at 15:49
From $P_1+P_0=1$ you get $P_1=1-P_0$. Hence put this into the first equation and solve it for $P_0$. Then you get $P_1$. – user8 Jan 17 '13 at 15:49
@ Azmoti its (4/5)P0 – Mat Lau. Jan 17 '13 at 15:59
@ n.c. thanks , i am trying it this way now , but it doesn't get me anywhere :-/ – Mat Lau. Jan 17 '13 at 16:00
bump... anyone ? – Mat Lau. Jan 17 '13 at 16:42

## 1 Answer

So let's do this:

We have $P_1=1-P_0$ as mentioned in my comment. Then plug this into the first equation leads to

$$P_0=\frac{4}{5}P_0+\frac{3}{4}(1-P_0)$$

which is equivalent to $$P_0-\frac{4}{5}P_0+\frac{3}{4}P_0=\frac{3}{4}$$

Write the left hand side as

$$P_0(1-\frac{4}{5}+\frac{3}{4})=\frac{3}{4}$$ now, $1-\frac{4}{5}+\frac{3}{4}=\frac{20-16+15}{20}=\frac{19}{20}$, so we have $$P_0\frac{19}{20}=\frac{3}{4}$$ therefore we end up with $$P_0=\frac{20\cdot 3}{4\cdot 19}=\frac{5\cdot 3}{19}=\frac{15}{19}$$

Going back to $P_1=1-P_0$, give $P_1=1-\frac{15}{19}=\frac{4}{19}$.

By the way, to ping someone in the comment, like @n.c. don't use a "space" between @ and the name. Otherwise the person will not be notified.

-
Thank you very much. Be blessed. – Mat Lau. Jan 17 '13 at 18:09
@MatLau You are welcome – user8 Jan 17 '13 at 18:10