Question

Out of 40 consecutive natural numbers, two are chosen at random. Probability that the sum of the numbers is odd, is
A) \[\dfrac{{14}}{{29}}\]
B) \[\dfrac{{20}}{{39}}\]
C) \[\dfrac{1}{2}\]
D) None of these

Answer 1

Hint: To solve this problem we should know that the sum of two odd numbers is always even. Again the sum of two even numbers is always even. So, to get the sum of 2 integers as odd one must be even and the other one must be odd. Then calculate all the sample space and required outcome using the concept of permutation and combination. And to calculate the required probability following detailed steps are done.

Complete step by step answer:
The sum of two odd numbers is always even. Again the sum of two even numbers is always even. So, to get the sum of 2 integers as odd one must be even and the other one must be odd.
Given the natural numbers are 1, 2,3,4,5,6,………,38,39,40 .
In 40 consecutive integers, 20 are even and 20 are odd.
The even natural numbers of 40 consecutive natural number are 2,4,6,…,36,38,40
The even natural numbers of 40 consecutive natural number are 1,3,5,…..,35,37,39
Now, the total outcome will be equal to select two random numbers from 40 consecutive natural number,
\[ \Rightarrow {40_{{C_2}}}\]
It is a form of combination,
\[ \Rightarrow {n_{{{\text{C}}_r}}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Now simplifying it,
\[ = \dfrac{{40!}}{{2!\left( {40 - 2} \right)!}}\]
\[ = \dfrac{{40 \times 39 \times 38!}}{{1 \times 2 \times \left( {38} \right)!}}\]
As, \[38!\] is in both numerator and denominator so it gets cancel out .Hence,
\[ = \dfrac{{40 \times 39}}{2}\]
\[ = 20 \times 39\]
\[ = 780\]
Similarly, to select one odd number from 20 natural numbers from the 40 consecutive natural number =\[{20_{{C_1}}}\]
Now simplifying it,
\[ \Rightarrow {20_{{{\text{C}}_1}}} = \dfrac{{20!}}{{1!\left( {20 - 1} \right)!}}\]
\[ = \dfrac{{20 \times 19!}}{{1 \times \left( {19} \right)!}}\]
As, \[19!\] is in both numerator and denominator so it gets cancel out .Hence,
\[ \Rightarrow {20_{{{\text{C}}_1}}} = 20\]
Similarly, to select one even number from 20 natural numbers from the 40 consecutive natural number =\[{20_{{C_1}}}\]
Now simplifying it,
\[ \Rightarrow {20_{{{\text{C}}_1}}} = \dfrac{{20!}}{{1!\left( {20 - 1} \right)!}}\]
\[ = \dfrac{{20 \times 19!}}{{1 \times \left( {19} \right)!}}\]
As, \[19!\] is in both numerator and denominator so it gets cancel out .Hence,
\[ \Rightarrow {20_{{{\text{C}}_1}}} = 20\]
Now we know that the probability formula of an event as,
Probability, \[ P\left( E \right) = \dfrac{{total{\text{ number of possible outcomes }}}}{{the{\text{ total number of outcomes}}}}\]
Therefore the required outcome is = \[{20_{{{\text{C}}_1}}} \times {20_{{{\text{C}}_1}}}\]
Therefore, probability that the sum of the numbers is odd, is
\[ = \dfrac{{{{20}_{{C_1}}} \times {{20}_{{C_1}}}}}{{{{40}_{{C_2}}}}}\]
\[ = \dfrac{{20 \times 20}}{{780}}\]
\[ = \dfrac{{400}}{{780}}\]
\[ = \dfrac{{20}}{{39}}\]
Hence the correct option is (B) \[\dfrac{{20}}{{39}}\].

Note:
To calculate such type of question, we should have knowledge about the permutation and combination and the basic concepts and formula of probability. We should also have basic concepts of natural numbers, real numbers etc. We should know that the sum of two odd numbers is always even. Again the sum of two even numbers is always even. So, to get the sum of 2 integers as odd one must be even and the other one must be odd.