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.