Out of $40$ consecutive integers two integers are drawn at random. The probability that their sum is odd is

To find probability of selecting two numbers out of $40$ integers

Total number of elements in sample-space$=$Number of ways of drawing two number out of $40$numbers$={}^{40}C_2 \left\{{}^{n}C_r=\frac{ n!}{ r! n-r!}\right\}$

$=\frac{ 40!}{2! (40-2!)}$

$=\frac{40!}{2(38!)}$

$=20\times 39$

If we are drawing two random integers whose sum is odd then it is possible only if one number is even and other is odd.

Since we know that there are $20$ even integers so the total cases of choosing one even integer$={}^{20}C_1=\frac{ 20!}{ 1! 19!}=20$

Now total odd number are $20$ so choosing one odd integer is$={}^{20}C_1=\frac{ 20!}{ 1! 19!}=20$

Probability$=\frac{\mathrm{number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes} }{\mathrm{total} \mathrm{outcomes}}$

Probability $=\frac{{}^{20}C_1\times {}^{20}C_1}{{}^{40}C_2}$

$=\frac{20\times 20}{20\times 39}=\frac{20}{39}$