I am solving a problem given as
Suppose we know there is a $60\%$ chance that it will rain tomorrow and a $70\%$
chance the high temperature will be above $30^o C$. Suppose we also know that there is a $40\%$ chance
that the high temperature will be above $30^oC$ and it will rain. How likely is it tomorrow will be a
dry day that does not go above $30^oC$?
The solution is given as –
We have one event $E$ which represents “It
will rain tomorrow” and another $F$ which represents “The high will be above $30^oC$ tomorrow”.
Our given probabilities tell us $P(E) = 0.6,\;P(F) = 0.7,$ and $P(E \cap F) = 0.4$. We are trying to
determine $P(E^c\cap F^c)$. As $E^c\cap F^c = (E \cup F)^c$.
$P(E \cup F) = P(E)+P(F)−P(E \cap F) = 0.7+0.6−0.4 = 0.9.$.
(This is the probability that it either will rain or be above $30^oC$).
Thus, $P(E^c \cap F^c) = P((E\cup F)^c) = 1 − P(E \cup F) = 1−0.9 = 0.1$
So there is a $10\%$ chance tomorrow will be a dry day that doesn’t reach $30$ degrees.
I am trying to solve this pictorially.
Our sample space is
Figure-1
The events $E$=it will rain tommorow, $F$= temperature will be greater than $30^o C$, and $E\cap F$ = it will rain tommorow and temperature will be greater than $30^o C$ are shown as
Figure-2
The event dry day and temperature doesn't go above $30$ degrees is shown as
Figure-3
From this figures, I have two doubts
i) From Figure-1, $E\cup F=S$. So, $P(E\cup F)=P(S)=1$.
But by formula $P(E\cup F)=P(E)+P(F)-P(E\cap F)=0.6+0.7-0.4=0.9\neq 1$
Why this is so?
ii) From Figure-3,
$P$(dry day and temperature below $30^o C$)=$P(E\cap F)^c=1-0.4=0.6\neq 0.1$
I have a doubt that why I get different answers.
I am not able to figure out where am I wrong?
Please clarify the doubt.
Best Answer
Your graphs do not indicate the venn diagrams correctly. Specifically, there are 4 disjoint events:$\{(R,HT),(R,\overline{HT}),(\overline{R},HT),(\overline{R},\overline{HT})\}$. the four parts of your rectangles should indicate these 4 disjoint events.
Which parts of the rectangle are covered by each event? See below:
rain $\equiv {(R,HT),(R,\overline{HT})}$
no rain $\equiv {(\overline{R},HT),(\overline{R},\overline{HT})}$
high temp $\equiv {(R,HT),(\overline{R},HT)}$
no high temp $\equiv {(R,\overline{HT}),(\overline{R},\overline{HT})}$
rain and high temp $\equiv {(R,HT)}$ no rain and no high temp $\equiv {(\overline R,\overline{HT})}$
Now coming to solving the question, the solution for which you already have, but see if the below helps:
Notations: $\mathbb P[R]$ = probability of rain tomorrow, $\mathbb P[HT]$ = probability of high temperature tomorrow, $\mathbb P[HT\cap R]$ = probability of rain with high temperature tomorrow. Further, $\mathbb P[\overline R]$ = probability of no rain tomorrow, $\mathbb P[\overline{HT}]$ = probability of low temperature.
From the laws of probability we know, 1) $\mathbb P[HT\cup R] = \mathbb P[HT]-\mathbb P[HT\cap R]+\mathbb P[R]$.
From De Morgan's laws, 2) $\mathbb P[\overline HT\cap \overline R] = \mathbb P[\overline {HT\cup R}] = 1- \mathbb P[HT\cup R]$
Now do you have all the ingredients to solve the question?