Question

Let A and B are two events such that the probability that exactly one of them occurs is $\dfrac{2}{5}$ and the probability that A or B occurs is $\dfrac{1}{2}$. Find the probability of both of them occurring together.
A. 0.01
B. 0.10
C. 0.20
D. 0.02

Answer 1

Hint: We first define the events and describe them in both set theory forms and Venn diagram. The individual events are independent of each other. We use the given values and use them in the formula $p\left( A\cup B \right)=p\left( A\cap {{B}^{c}} \right)+p\left( B\cap {{A}^{c}} \right)+p\left( A\cap B \right)$ to find the solution of the problem.

Complete step-by-step answer:
There are two events A and B. it’s given that the probability that exactly one of them occurs is $\dfrac{2}{5}$ and the probability that A or B occurs is $\dfrac{1}{2}$.
We try to express the given conditions in the form of Venn’s diagram and then in set theory.

The notion of one of them occurring is the sum of the probabilities of A happening and B happening. In set theory we express it as $p\left( A\cap {{B}^{c}} \right)+p\left( B\cap {{A}^{c}} \right)$. Here ${{A}^{c}},{{B}^{c}}$ defines the complement set of A and B respectively.
The notion of A or B occurring is the probability of either A happening or B happening. In set theory we express it as $p\left( A\cup B \right)$.
The total set described is the universal set. The event containing both A and B happening at least one is defined as $p\left( A\cup B \right)$.
The notion of A and B occurring together is the probability of A and B happening at the same time. In set theory we express it as $p\left( A\cap B \right)$.
So, $p\left( A\cap {{B}^{c}} \right)+p\left( B\cap {{A}^{c}} \right)=\dfrac{2}{5}$ and $p\left( A\cup B \right)=\dfrac{1}{2}$.
In set theory we have the formula $p\left( A\cup B \right)=p\left( A\cap {{B}^{c}} \right)+p\left( B\cap {{A}^{c}} \right)+p\left( A\cap B \right)$.
Putting the values, we get $\dfrac{1}{2}=\dfrac{2}{5}+p\left( A\cap B \right)$.
This gives us $p\left( A\cap B \right)=\dfrac{1}{2}-\dfrac{2}{5}=\dfrac{5-4}{10}=\dfrac{1}{10}=0.1$.
The probability of both of them occurring together is 0.1. The correct option is B.

So, the correct answer is “Option B”.

Note: The difference between $p\left( A\cap {{B}^{c}} \right)+p\left( B\cap {{A}^{c}} \right)$ and $p\left( A \right)+p\left( B \right)$ is that the later one describes the whole event where the intersection part of both events are also considered in each cases. But in the first case we are only taking the events happening in solo form.