# Problem in finding probability

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.

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?

• thanks for the answer. I have understood your answer. With the venn diagram you use, it is easy to understand the final answer. But may you please tell why my venn diagram is incorrect? I am not able to figure out the reason for that.
– Iti
Apr 11, 2021 at 7:42
• For me the main difficulty (in your diagram) is that the parts of the venn diagram should be disjoint. In your case, the four parts of the rectangle aren't disjoint Apr 11, 2021 at 7:45