Question

If $P\left( A \right)=\dfrac{1}{4}$, $P\left( B \right)=\dfrac{2}{5}$ and $P\left( A\cup B \right)=\dfrac{1}{2}$, find the value of the following:
a. $P\left( A\cap B \right)$
b. $P\left( A\cap B' \right)$
c. $P\left( A'\cap B \right)$
d. $P\left( A'\cup B' \right)$
e. $P\left( A'\cap B' \right)$

Answer 1

Hint : The value of the given expressions is calculated using theorems of probability. The addition and multiplication theorems of probability are used to obtain some important results which can be used as a direct formula for calculating the values of the given expressions.
$P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$
$P\left( A\cap B' \right)=P\left( A \right)-P\left( A\cap B \right)$
$P\left( A'\cap B \right)=P\left( B \right)-P\left( A\cap B \right)$
$P\left( A'\cup B' \right)=1-P\left( A\cap B \right)$
$P\left( A\cup B \right)=1-P\left( A'\cap B' \right)$

Complete step-by-step answer :
When two events cannot occur simultaneously, they are called mutually exclusive events.
If A and B were mutually exclusive events, $P\left( A\cap B \right)$ would have been zero. Therefore, $P\left( A\cup B \right)$ would have been calculated as:
$P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)$

When two events are independent, that is, the occurrence of one event does not affect the occurrence or non-occurrence of the other event, then,
$P\left( A\cap B \right)$ can be calculated as:
$P\left( A\cap B \right)=P\left( A \right)\cdot P\left( B \right)$

Given, $P\left( A \right)=\dfrac{1}{4}$, $P\left( B \right)=\dfrac{2}{5}$ and $P\left( A\cup B \right)=\dfrac{1}{2}$, therefore, $P\left( A\cap B \right)$ can be calculated as:
$P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$
$P\left( A\cap B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cup B \right)$
$P\left( A\cap B \right)=\dfrac{1}{4}+\dfrac{2}{5}-\dfrac{1}{2}$
$P\left( A\cap B \right)=\dfrac{5+8-10}{20}$
$P\left( A\cap B \right)=\dfrac{3}{20}$

$P\left( A\cap B' \right)$ can be calculated as:
$P\left( A\cap B' \right)=P\left( A \right)-P\left( A\cap B \right)$
$P\left( A\cap B' \right)=\dfrac{1}{4}-\dfrac{3}{20}$
$P\left( A\cap B' \right)=\dfrac{5-3}{20}$
$P\left( A\cap B' \right)=\dfrac{2}{20}$
$P\left( A\cap B' \right)=\dfrac{1}{10}$

$P\left( A'\cap B \right)$ can be calculated as:
$P\left( A'\cap B \right)=P\left( B \right)-P\left( A\cap B \right)$
$P\left( A'\cap B \right)=\dfrac{2}{5}-\dfrac{3}{20}$
$P\left( A'\cap B \right)=\dfrac{8-3}{20}$
$P\left( A'\cap B \right)=\dfrac{5}{20}$
$P\left( A'\cap B \right)=\dfrac{1}{4}$

$P\left( A'\cup B' \right)$is calculated as:
$P\left( A'\cup B' \right)=1-P\left( A\cap B \right)$
$P\left( A'\cup B' \right)=1-\dfrac{3}{20}$
$P\left( A'\cup B' \right)=\dfrac{17}{20}$

$P\left( A'\cap B' \right)$ is calculated as:
$P\left( A\cup B \right)=1-P\left( A'\cap B' \right)$
$P\left( A'\cap B' \right)=1-P\left( A\cup B \right)$
$P\left( A'\cap B' \right)=1-\dfrac{1}{2}$
$P\left( A'\cap B' \right)=\dfrac{1}{2}$

Note : The value of $P\left( A'\cap B' \right)$ and $P\left( A'\cup B' \right)$ is verified using $P\left( A'\cup B' \right)=P\left( A' \right)+P\left( B' \right)-P\left( A'\cap B' \right)$ as follows:
$\left( A'\cup B' \right)=P\left( A' \right)+P\left( B' \right)-P\left( A'\cap B' \right)$
$\dfrac{17}{20}=\dfrac{3}{4}+\dfrac{3}{5}-\dfrac{1}{2}$
$\dfrac{17}{20}=\dfrac{15+12-10}{20}$
$\dfrac{17}{20}=\dfrac{17}{20}$

Since, the term on the left hand side is equal to the term on the right hand side, therefore, the calculated values are correct.
We can also check the values of $P\left( A'\cap B \right)$ and $P\left( A\cap B' \right)$using the following identity,
$\left( A\cap B' \right)+P\left( A'\cap B \right)=P\left( A\cup B \right)-P\left( A\cap B \right)$
$\dfrac{1}{10}+\dfrac{1}{4}=\dfrac{1}{2}-\dfrac{3}{20}$
$\dfrac{2+5}{20}=\dfrac{10-3}{20}$
$\dfrac{7}{20}=\dfrac{7}{20}$

Since, the term on the left hand side is equal to the term on the right hand side, therefore, the calculated values are correct.
$P\left( A\cup B \right)$ can also be written as $P\left( A+B \right)$ and $P\left( A\cap B \right)$ can be written as $P\left( AB \right)$