Show that:
\[ \boldsymbol{ P(A) \ = \ \frac{ 1 \ – \ b \ – \ a }{ 1 \ – \ b } }\]
The aim of this question is to develop understanding of some of the basic probability and set theory properties for deriving some complex mathematical equations.
Expert Answer
Step 1: Given that:
\[ P(B) \ = \ b \]
And:
\[ P( \ \overline{A} \ \cap \ \overline{B} \ ) \ = \ a \]
Step 2: Since $A$ and $B$ are independent:
\[ P( \ A \ \cap \ B) \ = \ P(A)P(B) \]
Step 3: Deriving the required expression:
\[ P( \ \overline{A} \ \cap \ \overline{B} \ ) \ = \ a \]
Substituting the equation $\ \overline{A} \ \cap \ \overline{B} \ = \ \overline{A \ \cup \ B}$ in above expression:
\[ P( \ \overline{A \ \cup \ B} \ ) \ = \ a \ \]
Substituting the equation $ \ \overline{A \ \cup \ B} \ = \ 1\ \ – \ P( \ A \ \cup \ B \ )$ in above expression:
\[ 1 \ – \ P( \ A \ \cup \ B \ ) \ = \ a\]
Substituting the equation $ \ P( \ A \ \cup \ B \ )\ =\ P(A) \ + \ P(B) \ – \ P(A \cap B) $ in above expression:
\[ 1 \ – \ \{ \ P(A) \ + \ P(B) \ – \ P(A \cap B) \ \} \ = \ a \]
\[ 1 \ – \ \ P(A) \ – \ P(B) \ + \ P(A \cap B) \ = \ a \]
Substituting the equation $ P( \ A \ \cap \ B) \ = \ P(A) \cdot P(B) $ in above expression:
\[ 1 \ – \ \ P(A) \ – \ P(B) \ + \ P(A) \cdot P(B) \ = \ a \]
Substituting the equation $ P(B) \ = \ b $ in above expression:
\[ 1 \ – \ \ P(A) \ – \ b \ + \ P(A) \cdot b \ = \ a \]
Rearranging:
\[ 1 \ – \ a \ – \ b \ = \ P(A) \ – \ P(A) \cdot b\]
\[ 1 \ – \ a \ – \ b \ = \ P(A) \ ( \ 1 \ – \ b \ )\]
Rearranging:
\[ P(A) \ = \ \dfrac{ 1 \ – \ a \ – \ b }{ 1 \ – \ b } \]
Numerical Result
If $a$ is the joint probability of $A$ and $B$ not happening simultaneously and $b$ is the probability of $B$, then:
\[ P(A) \ = \ \dfrac{ 1 \ – \ a \ – \ b }{ 1 \ – \ b } \]
Example
If the joint probability of $A$ and $B$ not happening simultaneously is $0.2$ and the probability of $B$ is $0.1$, then find the probability of $A$.
From the above derivation:
\[ P(A) \ = \ \dfrac{ 1 \ – \ a \ – \ b }{ 1 \ – \ b } \]
\[ P(A) \ = \ \dfrac{ 1 \ – \ 0.2 \ – \ 0.1 }{ 1 \ – \ 0.1 } \]
\[ P(A) \ = \ \dfrac{ 0.7 }{ 0.9 } \]
\[ P(A) \ = \ 0.778 \]