Question

A and B are two events. Odds against A and 2:1. Odds in favor of $A \cup B$ are 3:1. If $x \le P\left( B \right) \le y$ , then the ordered pair (x, y) is
A. $\left( {\dfrac{5}{{12}},\dfrac{3}{4}} \right)$
B. $\left( {\dfrac{2}{3},\dfrac{3}{4}} \right)$
C. $\left( {\dfrac{1}{3},\dfrac{3}{4}} \right)$
D.None of these

Answer 1

Hint : From the first condition which is given in the question, we find the value of $P\left( A \right)$ (that is, the probability of findings A). And from the second condition, the value of $P\left( {A \cup B} \right)$ is found. With the help of the relation between $P\left( {A \cup B} \right)$ and $P\left( A \right)$ , we will try to find the range of $P\left( B \right)$ which is the required result.

Complete step-by-step answer :
We know that for any event,
$P\left( {Event} \right) = \dfrac{a}{{a + b}}$ , if its favor ratio is
And $P\left( {Event} \right) = 1 - \dfrac{a}{{a + b}}$ , if odd against is
Thus, we can write as follows:
$\Rightarrow P\left( A \right) = 1 - \dfrac{2}{{2 + 1}}$ , by just substituting
By simplifying, we get the value of as –
$\Rightarrow P\left( A \right) = \dfrac{1}{3}$
Similarly, we can write $P\left( {A \cup B} \right)$ as an event whose favor ratio is 3:1.
Thus, we get the equation as –
$\begin{array}{l} \Rightarrow P\left( {A \cup B} \right) = \dfrac{3}{{3 + 1}}\\\ \Rightarrow P\left( {A \cup B} \right) = \dfrac{3}{4} \end{array}$
Now, we know a relation –
$\Rightarrow P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$
But $P\left( {A \cap B} \right)$ is not given to us but however, it is greater than or equal to zero.
That means we are subtracting something from the sum of $P\left( A \right)$ and $P\left( B \right)$ to get $P\left( {A \cup B} \right)$ .
So we get an inequality as –
$\Rightarrow P\left( {A \cup B} \right) \le P\left( A \right) + P\left( B \right)$
But substituting the value of $P\left( {A \cup B} \right) = \dfrac{3}{4}$ and $P\left( A \right) = \dfrac{1}{3}$ , we get –
$\dfrac{3}{4} \le \dfrac{1}{3} + P\left( B \right)$
Subtracting $\dfrac{1}{3}$ on both sides, we get –
$\dfrac{3}{4} - \dfrac{1}{3} \le \dfrac{1}{3} + P\left( B \right) - \dfrac{1}{3}$
By simplifying, we get –
$P\left( B \right) \ge \dfrac{5}{{12}}$
And the maximum value of $P\left( B \right)$ is nothing but its union.
So,
$\begin{array}{l} \Rightarrow P\left( B \right) \le P\left( {A \cup B} \right)\\\ \Rightarrow P\left( B \right) \le \dfrac{3}{4} \end{array}$
Thus, we get the range of $P\left( B \right)$ as,
$\dfrac{5}{{12}} \le P\left( B \right) \le \dfrac{3}{4}$

So, the correct answer is “Option A”.

Note : The formula of odds and favor are very important in probability. Also students must analyze and understand why we have neglected the intersection in the union relation and why we have taken the maximum value of $P\left( B \right)$ to be $P\left( {A \cup B} \right)$ in order to get the required answer.