Question

There are two events A and B. If odds against A are as 2:1 and those in favor of \[A\cup B\] are as 3:1, then
(a) \[\dfrac{1}{2}\le P\left( B \right)\le \dfrac{3}{4}\]
(b) \[\dfrac{5}{12}\le P\left( B \right)\le \dfrac{3}{4}\]
(c) \[\dfrac{1}{4}\le P\left( B \right)\le \dfrac{3}{5}\]
(d) None of these

Answer 1

Hint: From the first condition, find the value of P(A). Now, from the second condition, find the value of \[P\left( A\cup B \right)\]. By the relation of \[P\left( A\cup B \right)\] with P(A) and P(B), try to find the minimum value of P(B). From using these conditions, the value of P(B) is nothing but its union. From this, find the range of P(B) which is the required result.

Complete step-by-step answer:
We are given the question that The odds against A is 2:1 and the favor of event \[A\cup B\] is 3:1. From odds against, we can write P(A) as below method:
If odds against is a:b, then,
\[P\left( A \right)=1-\dfrac{a}{a+b}\]
By substituting a = 2, b = 1, we get the equation as
\[P\left( A \right)=1-\dfrac{2}{2+1}\]
By simplifying, we get the equation in the form of
\[P\left( A \right)=1-\dfrac{2}{3}\]
By simplifying the above equation, we get the value of P(A) as:
\[P\left( A \right)=\dfrac{1}{3}....\left( i \right)\]
We can write \[P\left( A\cup B \right)\] by the following method.
If in favor ratio is c:d, then,
\[P\left( \text{Event} \right)=\dfrac{c}{c+d}\]
By substituting c = 3, d = 1, we get the equation as
\[P\left( E \right)=\dfrac{3}{3+1}\]
By simplifying, we can write the above equation as
\[P\left( A\cup B \right)=\dfrac{3}{4}....\left( ii \right)\]
We know that the relation of the union, the intersection is
\[P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)\]
We know that the intersection is greater than or equal to 0. So, we are subtracting a number from the sum. So, we can say an inequality given by the following:
\[P\left( A\cup B \right)\le P\left( A \right)+P\left( B \right)\]
By substituting the known values, we can write it as
\[\dfrac{3}{4}\le \dfrac{1}{3}+P\left( B \right)\]
By subtracting \[\dfrac{1}{3}\] on both the sides, we get it as
\[\dfrac{3}{4}-\dfrac{1}{3}\le \dfrac{1}{3}-\dfrac{1}{3}+P\left( B \right)\]
By simplifying, we can write the equation as
\[P\left( B \right)\ge \dfrac{5}{12}\]
The maximum value of P(B) is nothing but its union.
\[P(B)\le P\left( A\cup B \right)\]
By substituting the values, we can write it as
\[\dfrac{5}{12}\le P\left( B \right)\le \dfrac{3}{4}\]
Therefore, option (b) is the right answer.

Note: Be careful while using the union relation as you have to neglect the intersection to say the sum is greater than the union. The formula of odds and favor are very important in probability. Apply them carefully. The idea of the maximum value of P(B) is crucial to get the answer.