**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:

\[ P(A) \ = \ \dfrac{ 1 \ – \ a \ – \ b }{ 1 \ – \ b } \]

\[ P(A) \ = \ \dfrac{ 1 \ – \ 0.2 \ – \ 0.1 }{ 1 \ – \ 0.1 } \]

\[ P(A) \ = \ \dfrac{ 0.7 }{ 0.9 } \]

**\[ P(A) \ = \ 0.778 \]**