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**Hint:**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{align}

& P(A\cup B)=\dfrac{1}{2}+\dfrac{2}{3}-\dfrac{1}{3} \\

& \Rightarrow P(A\cup B)=\dfrac{5}{6} \\

\end{align}$

**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 $P(A\cup B)=P\left( A \right)+P\left( B \right)+P\left( A\cap B \right)$ .

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