Question

For two events A and B, if \[P\left( A \right)=P\left( A|B \right)=\dfrac{1}{4}\] and \[P\left( B|A \right)=\dfrac{1}{2}\] , then
(A) A and B are independent
(B) A and B are mutually exclusive
(C) \[P\left( A'|B \right)=\dfrac{3}{4}\]
(D) \[P\left( B'|A' \right)=\dfrac{1}{2}\]

Answer 1

Hint: It is given that \[P\left( A \right)=P\left( A|B \right)=\dfrac{1}{4}\] and \[P\left( B|A \right)=\dfrac{1}{2}\] . Use the formula, \[P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}\] and get the value of \[P\left( A\cap B \right)\] in terms of \[P\left( A \right)\] and \[P\left( B \right)\] . We know the condition when the events X and Y are independent, \[P\left( X \right)P\left( Y \right)=P\left( X\cap Y \right)\] . Use this condition and check whether the events A and B are independent or not. Now, get the value of \[P\left( A\cap B \right)\] using the formula, \[P\left( A\cap B \right)=P\left( A \right)P\left( B|A \right)\]. We know the condition when the events X and Y are mutually exclusive, \[P\left( X\cap Y \right)=0\] . Now, check whether the events A and B are exclusive or not. Since A and B are independent events, so A’ and B’ are also independent events. We know that the summation of the probabilities of any event and its complimentary event is equal to 1. So, \[P\left( A \right)+P\left( A' \right)=1\] and \[P\left( B \right)+P\left( B' \right)=1\] . Now, get the value of \[P\left( A' \right)\] . Since A’ and B’ are independent events so, \[P\left( A'|B' \right)=P\left( A' \right)\] . Since A and B are independent events so, \[P\left( B \right)=P\left( B|A \right)=\dfrac{1}{2}\] . Now, use \[P\left( B \right)+P\left( B' \right)=1\] and get the value of \[P\left( B' \right)\] . Since A’ and B’ are independent events so, \[P\left( B'|A' \right)=P\left( B' \right)\] . Now, solve it further and choose all the correct options.

Complete step-by-step answer:
According to the question, it is given that
 \[P\left( A \right)=P\left( A|B \right)=\dfrac{1}{4}\] ………………………….(1)
\[P\left( B|A \right)=\dfrac{1}{2}\] ……………………………(2)
We know the formula, \[P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}\] ……………………………………..(3)
Now, putting the value of \[P\left( A|B \right)\] from equation (3), in equation (1), we get
\[\begin{align}
  & P\left( A \right)=P\left( A|B \right) \\
 & \Rightarrow P\left( A \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)} \\
\end{align}\]
\[\Rightarrow P\left( A \right)P\left( B \right)=P\left( A\cap B \right)\] ………………………………..(4)
We know the condition when the events X and Y are independent, \[P\left( X \right)P\left( Y \right)=P\left( X\cap Y \right)\] ………..…………………(5)
From equation (4) and equation (5), we can say that the events A and B are independent events …………………………….(6)
We also know the formula, \[P\left( A\cap B \right)=P\left( A \right)P\left( B|A \right)\] ……………………………..(7)
From equation (1) and equation (2), we have
\[P\left( A \right)=\dfrac{1}{4}\] ,
\[P\left( B|A \right)=\dfrac{1}{2}\] .
Now, putting \[P\left( A \right)=\dfrac{1}{4}\] and \[P\left( B|A \right)=\dfrac{1}{2}\] in the formula shown in equation (7), we get
\[\begin{align}
  & \Rightarrow P\left( A\cap B \right)=P\left( A \right)P\left( B|A \right) \\
 & \Rightarrow P\left( A\cap B \right)=\dfrac{1}{4}\times \dfrac{1}{2} \\
\end{align}\]
\[\Rightarrow P\left( A\cap B \right)=\dfrac{1}{8}\] …………………………..(8)
We know the condition when the events X and Y are mutually exclusive, \[P\left( X\cap Y \right)=0\] ……………………………..(9)
From equation (8), we have the value of \[P\left( A\cap B \right)\] .
Since the value of \[P\left( A\cap B \right)\] is \[\dfrac{1}{8}\] so, the events A and B are not mutually exclusive.
From equation (6), we have that the events A and B are independent events.
Since the events A and B are independent so, the compliments of the events A and B are also independent events.
We know that when two events X and Y are independent events then \[P\left( X|Y \right)=P\left( X \right)\] or \[P\left( Y|X \right)=P\left( Y \right)\]…………………….(10)
The events A and B are also independent events. So, the events A’ and B’ are also independents.
Using equation (10), we can say that\[P\left( A'|B' \right)\] is equal to \[P\left( A' \right)\] . So,
\[P\left( A'|B' \right)=P\left( A' \right)\] …………………………..(11)
We know that the summation of the probabilities of any event and its complimentary event is equal to 1. So, \[P\left( A \right)+P\left( A' \right)=1\] ………………………(12)
Now, from equation (1) and equation (12), we get
\[\begin{align}
  & \Rightarrow P\left( A \right)+P\left( A' \right)=1 \\
 & \Rightarrow \dfrac{1}{4}+P\left( A' \right)=1 \\
 & \Rightarrow P\left( A' \right)=1-\dfrac{1}{4} \\
\end{align}\]
\[\Rightarrow P\left( A' \right)=\dfrac{3}{4}\] ………………………………………..(13)
Now, from equation (11) and equation (13), we get
\[P\left( A'|B' \right)=P\left( A' \right)=\dfrac{3}{4}\] ………………………………….(14)
Since A and B are independent events so, using equation (8), we can say that \[P\left( B \right)=P\left( B|A \right)=\dfrac{1}{2}\] ………………………(15)
We know that the summation of the probabilities of any event and its complimentary event is equal to 1. So, \[P\left( B \right)+P\left( B' \right)=1\] ……………………..(16)
Now, from equation (15) and equation (16), we get
\[\begin{align}
  & \Rightarrow P\left( B \right)+P\left( B' \right)=1 \\
 & \Rightarrow \dfrac{1}{2}+P\left( B' \right)=1 \\
 & \Rightarrow P\left( B' \right)=1-\dfrac{1}{2} \\
\end{align}\]
\[\Rightarrow P\left( B' \right)=\dfrac{1}{2}\] ……………………….(17)
Since A’ and B’ are independent events so, \[P\left( B'|A' \right)=P\left( B' \right)\] ………………………(18)
Now, from equation (17) and equation (18), we get
\[P\left( B'|A' \right)=P\left( B' \right)=\dfrac{1}{2}\] ……………………………………(19)
From equation (14) and equation (19), we have
\[P\left( A'|B' \right)=P\left( A' \right)=\dfrac{3}{4}\]
\[P\left( B'|A' \right)=P\left( B' \right)=\dfrac{1}{2}\]
Also, the events A and B are independent events.
Hence, option (A), option (C), and option (D) are correct.

Note: In this question, one might get confused because there is no information for the events A’ and B’. But we know the property that the summation of the probabilities of any event and its complimentary event is equal to 1. So, \[P\left( A \right)+P\left( A' \right)=1\] and \[P\left( B \right)+P\left( B' \right)=1\] .