Question

If $P(A) = \dfrac{2}{5}$, $P(B) = \dfrac{3}{{10}}$ and $P(A \cap B) = \dfrac{1}{5}$, then $P({A'}|{B'}) \cdot P({B'}|{A'})$ is equal to
(A) $\dfrac{5}{6}$
(B) $\dfrac{5}{7}$
(C) $\dfrac{{25}}{{42}}$
(D) $1$

Answer 1

Hint : To find the value of $P({A'}|{B'}) \cdot P({B'}|{A'})$ we will first find the value of $P({A'}|{B'})$ and then $P({B'}|{A'})$ using conditional probability formulas and then simplifying it using De Morgan’s Law. By substituting the data given in the question we will find the values of $P({A'}|{B'})$ and $P({B'}|{A'})$. At last, we will multiply $P({A'}|{B'})$ and $P({B'}|{A'})$ to find the result.

Complete step-by-step answer :
Given, Probability of event $A$ as $P(A) = \dfrac{2}{5}$, Probability of event $B$ as $P(B) = \dfrac{3}{{10}}$ and Probability of event $A$ and $B$ $P(A \cap B) = \dfrac{1}{5}$.
By conditional probability we know that $P(A|B)$ is the probability of occurrence of event $A$ when event $B$ has already occurred.
We know that if $A$ and $B$ are not independent, then the probability of the intersection of $A$ and $B$ i.e., the probability that both$A$ and $B$ occurs is given by $P(A|B) = \dfrac{{P(A \cap B)}}{{P(B)}}$.
Therefore, by same definition, the given expression will be
$\Rightarrow P({A'}|{B'}) = \dfrac{{P({A'} \cap {B'})}}{{P({B'})}}$ and $P({B'}|{A'}) = \dfrac{{P({B'} \cap {A'})}}{{P({A'})}}$
By De Morgan’s Law we know that the complement of intersection of sets $A$ and $B$ is equal to the union of ${A'}$ and ${B'}$ i.e., ${\left( {A \cap B} \right)'} = {A'} \cup {B'}$ and also the complement of the union of sets $A$ and $B$ is equal to the intersection of ${A'}$ and ${B'}$ i.e., ${\left( {A \cup B} \right)'} = {A'} \cap {B'}$.
Using this, we can write
$\Rightarrow P({A'}|{B'}) = \dfrac{{P{{(A \cup B)}'}}}{{P({B'})}}$ and $P({B'}|{A'}) = \dfrac{{P{{(B \cup A)}'}}}{{P({A'})}}$
As the probability of the complement of an event is the probability of that event subtracted from universal. Therefore, we can write
$\Rightarrow P({A'}|{B'}) = \dfrac{{1 - P(A \cup B)}}{{1 - P(B)}}$ and $P({B'}|{A'}) = \dfrac{{1 - P(B \cup A)}}{{1 - P(A)}}$ $- - - (1)$
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Putting the given values of $P(A) = \dfrac{2}{5}$, $P(B) = \dfrac{3}{{10}}$ and $P(A \cap B) = \dfrac{1}{5}$ , we get
$\Rightarrow P(A \cup B) = \dfrac{2}{5} + \dfrac{3}{{10}} - \dfrac{1}{5}$
On solving we get
$\Rightarrow P(A \cup B) = \dfrac{1}{2}$
We know that $P(A \cup B) = P(B \cup A)$. Therefore, we get
$\Rightarrow P(A \cup B) = P(B \cup A) = \dfrac{1}{2}$
Putting the given values of $P(A) = \dfrac{2}{5}$, $P(B) = \dfrac{3}{{10}}$, $P(A \cup B)$ and $P(B \cup A)$ in $(1)$ we get
$\Rightarrow P({A'}|{B'}) = \dfrac{{1 - \dfrac{1}{2}}}{{1 - \dfrac{3}{{10}}}}$ and $P({B'}|{A'}) = \dfrac{{1 - \dfrac{1}{2}}}{{1 - \dfrac{2}{5}}}$
On simplification we get
$\Rightarrow P({A'}|{B'}) = \dfrac{5}{7}$ and $P({B'}|{A'}) = \dfrac{5}{6}$
Now, on multiplying $P({A'}|{B'})$ and $P({B'}|{A'})$, we get
$\Rightarrow P({A'}|{B'}) \cdot P({B'}|{A'}) = \dfrac{5}{7} \times \dfrac{5}{6}$
On simplification we get
$\Rightarrow P({A'}|{B'}) \cdot P({B'}|{A'}) = \dfrac{{25}}{{42}}$
Therefore, we get $P({A'}|{B'}) \cdot P({B'}|{A'}) = \dfrac{{25}}{{42}}$.
So, the correct answer is “Option C”.

Note : This problem can also be solved by using the rules of set theory. Also, here one thing to note is that we have used De Morgan’s Law because here $P(A)$ means probability of an event $A$ and this $A$ also follows the rules of set theory. Therefore, we can use the methods of transformation with probability also. Remember probability always lies between 0 and 1.