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# 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. $\dfrac{5}{3}$

Last updated date: 17th Jun 2024
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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.

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