Question

Out of $40$ consecutive natural numbers, two are chosen at random. Probability that the sum of the two numbers is odd is

• A. $\frac{14}{39}$
• B. $\frac{20}{39}$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

The correct option is B

$\frac{20}{39}$

Explanation for the correct option

Step 1 : Finding the probability

Finding the total number of cases

We know that the number of ways of choosing $r$ objects from a collection of $n$ different objects is ${}^{n}C_r$.

Here, in this question, a total of $40$ consecutive natural numbers are given. Out of which any two are chosen. So, the total number of cases will be $N={}^{40}C_2$.

Step 2 : Finding the number of favorable cases

Let $A$ denote the event that the sum of the two chosen numbers is odd. It can be possible only when one number is even and the other is odd.

Now, out of $40$ consecutive natural numbers, there are $20$ even numbers and $20$ odd numbers. So, the event $A$ will happen only when out of $2$ chosen number, one number is chosen from the $20$ even numbers and one number is chosen from the $20$ odd numbers.

So, the total number of favorable cases will be

$m\left(A\right)={}^{20}C_1\times {}^{20}C_1$.

The required probability will be:

$$\begin{gathered} P\left(A\right)=\frac{m\left(A\right)}{N} \\ =\frac{{}^{20}C_1\times {}^{20}C_1}{{}^{40}C_2} \\ =\frac{20\times 20}{{\displaystyle \frac{40\times 39}{2}}} \\ =\frac{20}{39} \end{gathered}$$

Hence, the correct answer is Option (B).