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Suppose that A and B are independent events such that the probability that neither occurs is a and the probability of B is b.

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:

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

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

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

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

5/5 - (18 votes)