Difference Events Are Measurable
Suppose that a probability space \(\left( \Omega,\mathcal{F},P\right) \)is given. That is, the family of events \(\mathcal{F}\subset2^{\Omega}\)satisfies the axioms of a measurable space \begin{align*}
& \left( M_{1}\right) \ \mathcal{F}\not =\phi\\
& \left( M_{2}\right) \ \forall A\in\mathcal{F}:A^{c}\in\mathcal{F}\\
& \left( M_{3}\right) \ \forall\left\{ A_{n}\right\} _{n\in\mathbb{N} }\subset\mathcal{F}:\bigcup_{n\in\mathbb{N} }A_{n}\in\mathcal{F}\end{align*}and the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) satisfies the axioms of probability \begin{align*}
& \left( P_{1}\right) \ \forall A\in\mathcal{F}:P\left( A\right) \geq0\\
& \left( P_{2}\right) \ P\left( \Omega\right) =1\\
& \left( P_{3}\right) \ \forall\text{mutually disjoint }\left\{
A_{n}\right\} _{n\in\mathbb{N} }\subset\mathcal{F}:P\left( \bigcup_{n\in\mathbb{N} }A_{n}\right) =\sum_{n\in\mathbb{N} }P\left( A_{n}\right) .
\end{align*}
Given a probability space \(\left( \Omega,\mathcal{F},P\right) \), what can be said about the probability of a difference event? Before answering this question, we first verify that difference events are measurable.
A\backslash B\in\mathcal{F},
\]holds. Thus, \(\mathcal{F}\) is closed under set difference.
Probability of a Difference Event
Since the family of events \(\mathcal{F}\) is closed under set difference, for arbitrary events \(A,B\in\mathcal{F}\), \[
A\backslash B\in\mathcal{F},
\]holds. Therefore, the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) assigns the probability \(P\left( A\backslash B\right) \) to this event. We have \[
P\left( A\backslash B\right) =P\left( A\right) -P\left( A\cap B\right) .
\]In particular, if \(B\subset A\), then \[
A\cap B=B,
\]and hence
\[
P\left( A\backslash B\right) =P\left( A\right) -P\left( B\right) .
\]
P\left( A\backslash B\right) =P\left( A\right) -P\left( A\cap B\right)
\]holds. In particular, if \(B\subset A\), then \[
P\left( A\backslash B\right) =P\left( A\right) -P\left( B\right)
\]holds.
P\left( A\right) & =0.8\\
P\left( B\right) & =0.3.
\end{align*}Then
\begin{align*}
P\left( A\backslash B\right) & =P\left( A\right) -P\left( B\right)
\quad\because B\subset A\\
& =0.8-0.3\\
& =0.5.
\end{align*}
A\cap B & \in\mathcal{F}\\
A\cup B & \in\mathcal{F}\end{align*}holds. Then
\begin{align*}
P\left( \left( A\cup B\right) \backslash\left( A\cap B\right) \right) &
=P\left( A\cup B\right) -P\left( \left( A\cap B\right) \cap\left( A\cup B\right) \right) \quad\because\text{probability of a difference event}\\
& =P\left( A\cup B\right) -P\left( A\cap B\right) \quad\because A\cap B\subset A\cup B
\end{align*}holds.
\left( A\backslash B\right) \cup\left( B\backslash A\right) .
\]Since \(A\backslash B\) and \(B\backslash A\) are mutually disjoint, \begin{align*}
P\left( \left( A\backslash B\right) \cup\left( B\backslash A\right)
\right) & =P\left( A\backslash B\right) +P\left( B\backslash A\right)
\quad\because\text{finite additivity}\\
& =\left[ P\left( A\right) -P\left( A\cap B\right) \right] +\left[
P\left( B\right) -P\left( A\cap B\right) \right] \quad\because
\text{probability of a difference event}\\
& =P\left( A\right) +P\left( B\right) -2P\left( A\cap B\right)
\end{align*}holds.
Range of Possible Values of the Probability of a Difference Event
The following result gives the range of possible values of the probability of a difference event.
P\left( A\right) & =a\\
P\left( B\right) & =b.
\end{align*}Then
\[
\max\left\{ 0,a-b\right\} \leq P\left( A\backslash B\right) \leq
\min\left\{ a,1-b\right\}
\]holds.
We now examine the meaning of the inequality
\[
\max\left\{ 0,a-b\right\} \leq P\left( A\backslash B\right) \leq
\min\left\{ a,1-b\right\}
\]obtained from the preceding proposition. The difference event \(A\backslash B\) is the event that \(A\) occurs while \(B\) does not occur. Therefore, if \(P\left( A\right) \) is greater than \(P\left( B\right) \), at least the difference between these probabilities must remain outside \(B\), so \[
P\left( A\backslash B\right) \geq a-b.
\]On the other hand, if \(a\leq b\), it is possible for all of \(A\) to be contained in \(B\), in which case \(P\left( A\backslash B\right) =0\). This explains the lower bound \[
\max\left\{ 0,a-b\right\} .
\]
For the upper bound, since \(A\backslash B\subset A\), \[
P\left( A\backslash B\right) \leq a.
\]Also, since \(A\backslash B\subset B^{c}\), \[
P\left( A\backslash B\right) \leq P\left( B^{c}\right) =1-b.
\]Therefore,
\[
P\left( A\backslash B\right) \leq\min\left\{ a,1-b\right\} .
\]
P\left( A\right) & =\frac{3}{5}\\
P\left( B\right) & =\frac{1}{2}.
\end{align*}Then, by the preceding proposition,
\begin{align*}
\max\left\{ 0,\frac{3}{5}-\frac{1}{2}\right\} & \leq P\left( A\backslash B\right) \\
& \leq\min\left\{ \frac{3}{5},1-\frac{1}{2}\right\} ,
\end{align*}and hence
\[
\frac{1}{10}\leq P\left( A\backslash B\right) \leq\frac{1}{2}.
\]
Exercises
\begin{align*}
P(A) & =\frac{7}{10}\\
P(B) & =\frac{1}{2}\\
P(A\cap B) & =\frac{3}{10}.
\end{align*}Find
\begin{align*}
& P(A\backslash B)\\
& P(B\backslash A).
\end{align*}
\[
B\subset A
\]and
\begin{align*}
P(A) & =\frac{4}{5}\\
P(B) & =\frac{3}{10}.
\end{align*}Find
\[
P(A\backslash B).
\]
P(A) & =\frac{3}{5}\\
P(B) & =\frac{1}{2}\\
P(A\cap B) & =\frac{1}{4}.
\end{align*}Find the probability that exactly one of \(A\) and \(B\) occurs.
P(A) & =\frac{2}{3}\\
P(A\backslash B) & =\frac{1}{4}.
\end{align*}Find
\[
P(A\cap B).
\]
\Omega=\left\{ \omega_{1},\omega_{2},\omega_{3},\omega_{4}\right\} ,
\]the event space is
\[
\mathcal{F}=2^{\Omega},
\]and the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) satisfies \begin{align*}
P\left( \left\{ \omega_{1}\right\} \right) & =\frac{1}{10}\\
P\left( \left\{ \omega_{1},\omega_{2}\right\} \right) & =\frac{1}{2}\\
P\left( \left\{ \omega_{1},\omega_{2},\omega_{3}\right\} \right) &
=\frac{7}{10}.
\end{align*}Find
\[
P\left( \left\{ \omega_{2},\omega_{4}\right\} \right) .
\]
\[
\Omega=\left\{ 1,2,3,4,5,6\right\} .
\]Let the event space be
\[
\mathcal{F}=2^{\Omega}\]and let the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) assign to each event \(A\in\mathcal{F}\) the probability \[
P\left( A\right) =\frac{\left\vert A\right\vert }{\left\vert \Omega
\right\vert }=\frac{\left\vert A\right\vert }{6}.
\]Then \(\left( \Omega,\mathcal{F},P\right) \) is a probability space. Find the probability of the event that an odd number is rolled but a number greater than or equal to \(4\) is not rolled. Use the formula for the probability of a difference event.
\[
\Omega=\left\{ \left( i,j\right) \ |\ i,j\in\left\{ 1,2,3,4,5,6\right\}
\right\} .
\]Let the event space be
\[
\mathcal{F}=2^{\Omega}\]and let the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) assign to each event \(A\in\mathcal{F}\) the probability \[
P\left( A\right) =\frac{\left\vert A\right\vert }{\left\vert \Omega
\right\vert }=\frac{\left\vert A\right\vert }{36}.
\]Then \(\left( \Omega,\mathcal{F},P\right) \) is a probability space. Find the probability of the event that the first roll is even but the second roll is not even. Use the formula for the probability of a difference event.
\begin{align*}
P(A) & =\frac{4}{5}\\
P(A\cap B) & =\frac{1}{2}\\
P(A\cap C) & =\frac{2}{5}\\
P(A\cap B\cap C) & =\frac{1}{4}.
\end{align*}Find
\[
P\left( A\backslash(B\cup C)\right) .
\]
