Question

What is the probability of the occurrence of a number that is odd or less than 5 when a fair die is rolled?
A. $\dfrac{5}{6}$
B. $\dfrac{1}{3}$
C. $\dfrac{2}{3}$
D. $\dfrac{5}{3}$

Answer 1

To find the probability of the occurrence of a number that is odd or less than 5 when a fair die is rolled, first write down the sample space. From this, select the odd numbers and let us denote this set to be A. Similarly, we will consider B to be the set of numbers less than 5. Now, find the probabilities of each of these. Occurrence of a number that is odd or less than 5 is denoted as $A\cup B$ . We can find the probability of this by using the formula $P(A\cup B)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$ . $A\cap B$ represents the set that has elements common in A and B. Finding $P\left( A\cap B \right)$ and substituting the values in the formula, the required value can be calculated.

Complete step by step answer:
We need to find the probability of the occurrence of a number that is odd or less than 5 when a fair die is rolled.
We know that when a die is thrown, the sample space will be given by
S=\left\\{ 1,2,3,4,5,6 \right\\}
Of these, the odd numbers are \left\\{ 1,3,5 \right\\}
Let us consider A to be the set of odd numbers occurring when the die is thrown. This is given as
A=\left\\{ 1,3,5 \right\\}...(i)
Let us consider B to be the numbers less than 5 and will be
B=\left\\{ 1,2,3,4 \right\\}...(ii)
Let us now calculate the probability of odd numbers.
$P(A)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
From S, we know that the total number of outcomes =6.
From A, number of favourable outcomes =3
Thus,
$P(A)=\dfrac{3}{6}=\dfrac{1}{2}...(iii)$
Now, let us now calculate the probability of odd numbers less than 5.
From B, number of favourable outcomes =2
$P(B)=\dfrac{4}{\text{6}}=\dfrac{2}{3}...(iv)$
We need to find the probability of occurrence of a number that is odd or less than 5. This is denoted as $P\left( A\cup B \right)$ .
We know that $P(A\cup B)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)...(a)$
$P\left( A\cap B \right)$ is the probability of occurrences common to both A and B.
We can see from (i) and (ii) that
A\cap B=\left\\{ 1,3 \right\\}
Now, we can find $P\left( A\cap B \right)$ .
$P\left( A\cap B \right)=\dfrac{2}{6}=\dfrac{1}{3}...(v)$
Let us substitute (iii), (iv) and (v) in (a). We get
$\begin{aligned} & P(A\cup B)=\dfrac{1}{2}+\dfrac{2}{3}-\dfrac{1}{3} \\\ & \Rightarrow P(A\cup B)=\dfrac{5}{6} \\\ \end{aligned}$

So, the correct answer is “Option A”.

Note: The representations must be carefully chosen. When we say A or B, we will be using $A\cup B$ . When we say A and B, we will be using $A\cap B$ . Also, you may make mistakes by writing the equation for $P(A\cup B)$ as $P(A\cup B)=P\left( A \right)+P\left( B \right)+P\left( A\cap B \right)$ .