I'm trying to understand what the sample (Kolmogorov, elementary event) space and event (Kolmogorov, random event) space is for the following problem. It has a solution which someone has presented, however I'm trying to put it in terms of above. I think some of my confusion if these are mutually exclusive events, $P(A) + P(B)$ cannot add up to more than $1$. Can anyone help here?
Suppose we have the following information:
- There is a 60 percent chance that it will rain today.
- There is a 50 percent chance that it will rain tomorrow.
- There is a 30 percent chance that it does not rain either day.
Find the following probabilities:
- The probability that it will rain today or tomorrow.
- The probability that it will rain today and tomorrow.
- The probability that it will rain today but not tomorrow.
- The probability that it either will rain today or tomorrow, but not both.
A is the event it will rain today and B is the event it will rain tomorrow.
- $P(A) = 0.6$
- $P(B) = 0.5$
- $P(A^c \cap B^c) = 0.3$
Best Answer
The sample space can be broken down into $4$ mutually exclusive events events $A\cap B^c,A^c\cap B,A\cap B,A^c\cap B^c$. Since $P(A)=0.6,P(A^c)=0.4$. Using this $P(A^c\cap B^c)=0.3$ leads to $P(A^c\cap B)=0.1$. Similarly $P(B)=0.5$ gives $P(B^c)=0.5$ and $P(A\cap B^c)=0.2$ Finally $P(A\cap B)=1-(0.3+0.1+0.2)=0.4$, since the 4 events described at the beginning make up the whole space.
Your four statements are: (Two ways to get first) $P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.5+0.6-0.4=0.7.$ $P(A\cup B)=P(A\cap B)+P(A\cap B^c)+P(A^c\cap B)= 0.4+0.2+0.1=0.7$.
$P(A\cap B)=0.4$
$P(A\cap B^c)=0.2$
$P(A\cap B^c)+P(A^c\cap B)=0.2+0.1=0.3$.