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.

  1. $P(A) = 0.6$
  2. $P(B) = 0.5$
  3. $P(A^c \cap B^c) = 0.3$

2 Answers 2


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) $P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.5+0.6-0.4=0.7.$ $P(A\cup B)=P(A\cap B)+P(A\cap B^c)+P(A^c\cap B)= 0.4+0.2+0.1=0.7$.

$P(A\cap B)=0.4$

$P(A\cap B^c)=0.2$

$P(A\cap B^c)+P(A^c\cap B)=0.2+0.1=0.3$.

  • $\begingroup$ I added a few missing claims which pertain to the information overload. In general I'm trying to understand what the sample space is. It's my understanding all the elementary events in the sample space have their probabilities add up to 1. The probability 6/10 and 5/10 are confusing me in this task. $\endgroup$
    – Nick
    Aug 30, 2020 at 21:11
  • $\begingroup$ @Nick I've changed my answer to reflect your revised question. $\endgroup$ Aug 31, 2020 at 2:20
  • $\begingroup$ This is helpful. "Using this $P(A^c \cap B^c) = 0.3$ leads to $P(A^c \cap B) = 0.1$. How did you arrive at that? $\endgroup$
    – Nick
    Aug 31, 2020 at 20:18
  • $\begingroup$ $P(A^c\cap B)+P(A^c\cap B^c)=P(A^c)=0.4$ $\endgroup$ Aug 31, 2020 at 22:00
  • $\begingroup$ Why does this equality hold? Is it because $P(A^c \cap B) + P(A^c \cap B^c)$ is $A^c \cap (B \cup B^c) = A^c \cap S = A^c$? $\endgroup$
    – Nick
    Aug 31, 2020 at 22:52

We know $$\Pr(A\cup B)+\Pr(A^c\cap B^c)=1$$ because the events are mutually exclusive and exhaustive.

We also know $$\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)$$

Can you continue from here?

  • $\begingroup$ I'm stuck because I'm trying to define the sample space and mutually exclusive events such that P(E) = 1. I'm getting confused by the 6/10 and 5/10 probabilities and how if these were mutually exclusive elementary events in the sample space, they would be <= 1. $\endgroup$
    – Nick
    Aug 30, 2020 at 21:18
  • 1
    $\begingroup$ There are four points in the sample space: It rains today and tomorrow; it rains today but not tomorrow; it rains tomorrow but not today; it rains neither day. $\endgroup$
    – saulspatz
    Aug 30, 2020 at 21:23
  • $\begingroup$ @saulspatz Im confused on why it dont have rain today or tomorrow? can you explain thanks! $\endgroup$
    – Remu X
    Jul 25 at 5:57
  • 1
    $\begingroup$ @RemuXu Rain today or tomorrow comprises $3$ of the points; all but $A^c\cap B^c$ $\endgroup$
    – saulspatz
    Jul 26 at 19:32

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