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.

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:

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

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

$P(A) \ = \ \dfrac{ 0.7 }{ 0.9 }$

$P(A) \ = \ 0.778$
The text passage contains a detailed derivation of a probability formula and an example that applies the derived formula to a specific case.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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.

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:

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

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

$P(A) \ = \ \dfrac{ 0.7 }{ 0.9 }$

$P(A) \ = \ 0.778$
user
The text passage contains a detailed derivation of a probability formula and an example that applies the derived formula to a specific case.

TYPE: SINGLE_QUESTION_WITH_ANSWER