Question

Prove that \[P(A - B) = P(A) - P(A \cap B) = P(A \cup B) - P(B) = P(A \cap \bar B) = 1 - P(\bar A \cup B)\]

Answer 1

Hint: We have here 5 things to prove equivalent to each other. We will prove 2 at a time. We will write P(A) in terms of A and B to get the first two, then use that result in the formula of $P(A \cup B)$ to get the equivalence of first and third. Now use the DE Morgan’s law to find equivalence between the third and fourth.

Complete step-by-step answer:
First of all we will prove that $P(A \cap \bar B) = P(A) - P(A \cap B)$.
We know that:- $A = A \cap U$, where U is the Universal set for any set $A \subseteq U$.
Therefore, we can write:- $P(A) = P(A \cap U)$
We also know that for any set $B \subseteq U,B \cup \bar B = U$, where $\bar B$ is the complement of B.
Therefore, $P(A) = P(A \cap U) = P\{ A \cap (B \cup \bar B)\} $
We will use the fact that $A \cap (B \cup \bar B) = (A \cap B) \cup (A \cap \bar B)$
So, we will get:- $P(A) = P\{ A \cap (B \cup \bar B)\} = P\{ (A \cap B) \cup (A \cap \bar B)\} $
Now using the formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
So, we will have:- $P(A) = P(A \cap B) + P(A \cap \bar B) - P\{ (A \cap B) \cap (A \cap \bar B)\} $ …….(1)
Let us consider $(A \cap B) \cap (A \cap \bar B)$.
We can write it as: \[(A \cap B) \cap (A \cap \bar B) = (A \cap A) \cap (B \cap \bar B) = A \cap \phi = \phi \]
(Because $B \cap \bar B = \phi $ always)
So, rewriting (1), we will get:- $P(A) = P(A \cap B) + P(A \cap \bar B) - P(\phi ) = P(A \cap B) + P(A \cap \bar B)$
(Because $P(\phi ) = 0$)
So on rearranging the terms, we will have:-
$P(A \cap \bar B) = P(A) - P(A \cap B)$ ………(2)
Now, we will show that \[P(A - B) = P(A \cup B) - P(B)\]
Now, consider the formula:- $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Rewriting it as:- $P(A \cup B) = \{ P(A) - P(A \cap B)\} + P(B)$
Using (2), we can rewrite it as:- $P(A \cup B) = P(A - B) + P(B)$
On rearranging the terms, we will get:-
$P(A - B) = P(A \cup B) - P(B)$ …………(3)
Now, we know that:- $P(A) = 1 - P(\bar A)$.
Using this method on \[P(A \cap \bar B)\], we get:- \[P(A \cap \bar B) = 1 - P\{ (\overline {A \cap \bar B} )\} \]
By De Morgan’s law, we know that:- \[(\overline {A \cap \bar B} ) = \bar A \cup \bar {\bar B} = \bar A \cup B\] (Because $\bar {\bar B} = B$ always)
So, we have:- \[P(A \cap \bar B) = 1 - P\{ (\overline {A \cap \bar B} )\} = 1 - P(\bar A \cup B)\] ………….(4)
At last, we will prove that \[P(A \cap \bar B) = P(A) - P(A \cap B)\]
Rewriting LHS as:- \[P(A \cap \bar B) = P(A \cap (U - B))\]
On simplifying the RHS, we will get:-
\[P(A \cap \bar B) = P(A \cap U) - P(A \cap B)\]
On simplifying it further, we get:-
\[P(A \cap \bar B) = P(A) - P(A \cap B)\] ……………(5)
Combining (2), (3), (4) and (5), we will get the desired result which is given by:-
\[P(A - B) = P(A) - P(A \cap B) = P(A \cup B) - P(B) = P(A \cap \bar B) = 1 - P(\bar A \cup B)\]

Note: The students may note that they may choose to prove any of them equal to any one irrespective of their order in the question because in the end eventually, we are proving all of them equal to each other.
The students, if confused between the validity of anything in this solution may use the Venn Diagram to see it for their satisfaction.