Question

If events A and B are independent and $P\left( A \right)=0.15,\,\,P\left( A\cup B \right)=0.45$, then $P\left( B \right)$ is:
(a) $\dfrac{6}{13}$
(b) $\dfrac{6}{17}$
(c) $\dfrac{6}{19}$
(d) $\dfrac{6}{23}$

Answer 1

Hint: Use the given condition to find the value of intersection of events A, B. Now draw Venn diagrams of two events X, Y. Then derive a formula for their union. Try to relate union, intersection and the two events themselves for this formula. Now substitute all the values you have. Then you will be left with a single variable $P\left( B \right)$. Now keep this $P\left( B \right)$ variable term on the left-hand side and send all other constants on to the right-hand side. By this, after all simplification the values on the right hand side will be our required result that is the value of $P\left( B \right)$.

Complete step-by-step answer:
Venn Diagram: A Venn diagram is a diagram that shows all possible logical relations between a collection of different sets. These diagrams depict elements as points in a space, and as set as regions inside a closed curve generally circles. Each circle represents a set. The overlapping region represents the common points between sets.
Intersection of sets: In mathematics, intersection of 2 sets A, B is the set consisting of all elements common in both A, B denoted by $A\cap B$.
Union of sets: In mathematics, union of 2 sets A, B is the set consisting of all elements belonging to A, B denoted by $A\cup B$.
Let us take any two random events X, Y.
Let us assume the probability of events X is P(X).
Let us assume the probability of event y is P(Y).
Let us assume intersection of events as $P\left( X\cap Y \right)$
Let us assume union of events as $P\left( X\cup Y \right)$

First Venn diagram is representation of X and the second Venn diagram is representation of the event y.
By the diagram, when we combine both X and Y events, we count the intersection part twice. So, we need to subtract the intersection term once to get the union. So, we can write it as:
$P\left( X\cup Y \right)=P\left( X \right)+P\left( Y \right)-P\left( X\cap Y \right)$
It is given in the question they are independent, so, we say:
$P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right)$
Given in the question the values of few terms, we get:
$P\left( A\cup B \right)=0.45,P\left( A \right)=0.15$
By substituting events A, B into union formula, we get:
$P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$
By substituting the value of $P\left( A\cap B \right)$ in the equation, we get:
$P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A \right)\times P\left( B \right)$
By substituting the values, you know, we get it as:
$0.45=0.15+P\left( B \right)-0.15P\left( B \right)$
So, simplifying, we get $P\left( B \right)$ value, as below:
$P\left( B \right)=\dfrac{30}{25}=\dfrac{6}{17}$
Therefore option (b) is the correct answer for the given question.

Note: Generally, students confuse and take $P\left( A\cap B \right)=0$ when they are all independent but it is a wrong method. You must always take $P\left( A\cap B \right)=P\left( A \right)+P\left( B \right)$. While substituting don’t forget the term $P\left( B \right)$ generated by intersection probability. Students forget that term and solve where they get the wrong result. So, solve it carefully.