I'm supposed to prove that if $A \subset B$, then $P(A) \leq P(B)$. The hint it gives is confusing me even more. It says use a venn diagram to convince yourself $ B = A \cup (A^c \cap B)$ and $A$ and $A^c \cap B$ are disjoint.

I know that I can do this:

$A \cup (A^c \cap B) = (A \cup A^c) \cap (A \cup B)$

Since $(A \cup A^c) = U$ that means if it intersects with $A \cup B$ we just have a venn diagram with everything shaded in.

How do I use this to my advantage to prove the original question? And wouldn't the two sets $A$ and $A^c \cap B$ being disjoint hurt me? Since I want to show that everything in $A$ is in $B$?

  • $\begingroup$ @bof - $P(A)$ denotes the probability measure of the set $A$. $\endgroup$
    – user169852
    Apr 5, 2015 at 5:28
  • $\begingroup$ That's kind of why I'm so confused about these proofs. I don't know what $P(B)$ and $P(A)$ are. And $P(A) \leq P(B)$ just means the probability of A happening is less than or equal to the probability of B happening. $\endgroup$
    – Alex
    Apr 5, 2015 at 5:28
  • $\begingroup$ I don't understand why anyone would downvote this question! $\endgroup$
    – user21820
    Apr 5, 2015 at 6:23

4 Answers 4


You probably learned a fact on the lines of "if two events $X$ and $Y$ are disjoint and independent, then $P(X\cup Y)=P(X)+P(Y)$." Since $A$ and $A^c\cap B$ are disjoint, you have \begin{align*} P(B)&=P(A\cup(A^c\cap B))\\&=P(A)+P(A^c\cap B)\\&\geq P(A)+0\\&=P(A) \end{align*} where we used the fact that $P(A^c\cap B)\geq 0$.

  • $\begingroup$ What is going on in the third line where we have $P(A) + 0$? $\endgroup$
    – Alex
    Apr 5, 2015 at 5:33
  • $\begingroup$ That's just using the fact that $P(A^c\cap B)\geq 0$, since probabilities can't be negative. $\endgroup$ Apr 5, 2015 at 5:43
  • $\begingroup$ Why is $A$ and $A^c \cap B$ disjoint? What does it look like in a venn diagram? $\endgroup$
    – CountDOOKU
    Mar 5, 2022 at 6:57
  • $\begingroup$ @CountDOOKU Try drawing the diagram, it's very straightforward visually. $\endgroup$ Mar 7, 2022 at 10:07

$B = A \cup (B\setminus A)$

$P(B) = P(A) + P(B\setminus A) - 0 $

Note: $A$ and $B\setminus A$ are disjoint.

Therefore, $P(A) \leq P(B)$

  • $\begingroup$ Answer is valid, even if it does not say anything about the confusing hint mentioned by @Alex. $\endgroup$
    – mlc
    Mar 16, 2017 at 17:11

This is my first post on math StackExchange. I was reading up on probabilities when I came across this question. The following is my shot at the problem. Do let me know in the comments if this is a valid proof.

$P(B) = P(A) + P(A^C \cap B)$

The above equation can be formulated by imagining a simple Venn diagram. Now, by removing the second probability from R.H.S., we get,

$P(B) \geq P(A)$


Observe that $B=(B\cap A)\cup (B\cap A^c)=A\cup (B\cap A^c)$.

Therefore $$P(B)=P(A\cup (B\cap A^c))=P(A)+P(B\cap A^c)-P(A\cap (B\cap A^c))\\ =P(A)+P(B\cap A^c)-P(\varnothing)\\=P(A)+P(B\cap A^c)-0\\=P(A)+P(B\cap A^c).$$

Since $P(B\cap A^c)\ge 0$ we have that $P(B)\ge P(A)$.

  • $\begingroup$ Isn't $A \cup (B \cap A^c) = (A \cup B) \cap (A \cup A^c)$? $\endgroup$
    – Alex
    Apr 5, 2015 at 5:54
  • $\begingroup$ I just apply $P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)$. Here $X=A$ and $Y=B\cap A^c$. $\endgroup$
    – ASB
    Apr 5, 2015 at 6:02

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