I have a question that I need to solve. It goes like this:
The chance of rain today if it was rainy yesterday is $0.7$. The chance of rain today if it was not rainy yesterday is $0.2$. We know that it was not rainy on Sunday. Given the fact that it was rainy on Tuesday, what are the chances that it was rainy on Monday?
I believe that it's $0.2(0.2\cdot0.7+0.7\cdot0.7)$ but I'm not sure.
On
Let $R_s$, $R_m$ and $R_t$ be the events that it was raining on sunday/monday/tuesday and $R_s^c$, $R_m^c$ and $R_t^c$ be the events that it was not raining on sunday/monday/tuesday.
We want to compute $\Pr (R_m \ \vert \ R_t)$ and in all our calculations we need to keep in mind that we are operating in a universe in which $R_s^c$ is a given fact. Applying Bayes theorem, we can compute:
$$\Pr (R_m \ \vert R_t) =\frac{ \Pr (R_t \ \vert R_m) \Pr (R_m)} {\Pr (R_t)}$$.
We observe:
$$ \Pr (R_t \ \vert R_m) = 0.7 $$ $$ \Pr (R_m) = 0.2 $$ $$ \Pr (R_t) = \Pr (R_t \ \vert R_m)\Pr (R_m) + \Pr (R_t \ \vert R_m^c)\Pr (R_m^c) = 0.7*0.2 + 0.2*0.8 = 0.3 $$
In total: $$\Pr (R_m \ \vert R_t) = 0.7*0.2 / 0.3 = (14/100)/(30/100) = 14/30 = 7/15 $$
$$\text{Sunday not rainy} \cases{ 0.2\text{: Monday rainy}\cases{ 0.7\text{: Tuesday rainy}\\ 0.3\text{: Tuesday not rainy} }\\ 0.8\text{: Monday not rainy}\cases{ 0.2\text{: Tuesday rainy}\\ 0.8\text{: Tuesday not rainy}} }$$
$$\begin{align*} \Pr (\text{Monday rainy} \mid \text{Tuesday rainy}) &=\frac{\Pr(\text{Monday rainy} \cap \text{Tuesday rainy})}{\Pr(\text{Tuesday rainy})}\\ &=\frac{ \Pr(\text{Monday rainy})\Pr(\text{Tuesday rainy}\mid\text{Monday rainy})}{\Pr(\text{Tuesday rainy})}\\ &= \frac{0.2\cdot 0.7}{0.2\cdot 0.7 + 0.8\cdot0.2}\\ &= \frac7{15} \end{align*}$$