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?
Best Answer
Use the fact that $$P(A\cup B) = P(A) + P(B) - P(A\cap B).$$ This gives you $$P(A\cup B) = 1.1 - P(A\cap B).$$ Since $P(A\cup B) \le 1$ we have $$1\ge 1.1 - P(A\cap B),$$ yielding $P(A\cap B)\ge .1$. If $A\subseteq B$, then $P(A\cap B) = P(A) = .4.$ This represents the largest possible value for $P(A\cap B)$. so $.1\le P(A\cap B) \le .4$